Integrals: Chapter 7 Links
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Integrals 7.5 focuses on the integration of rational functions, where both the numerator and denominator are polynomials. Such integrals are evaluated by first expressing the given function in a suitable form and then applying the method of partial fractions.
\displaystyle \int \frac{P(x)}{Q(x)}\,dx
If the degree of the numerator is greater than or equal to the degree of the denominator, the given expression is called an improper rational function. In such cases, we first convert it into a proper rational function by using long division. If possible, we also factorise the denominator before proceeding further.
Once the rational function is in proper form, we decompose it into partial fractions and integrate each simpler fraction separately using standard integration formulas. In this exercise, we will learn these techniques step by step and use them to solve different types of rational function integrals.
Key Concepts
1. Rational Functions
A rational function is a function of the form
\displaystyle \frac{P(x)}{Q(x)},\qquad Q(x)\neq 0
where P(x) and Q(x) are polynomials. In this exercise, we evaluate the integrals of rational functions by expressing them as simpler fractions using the method of partial fractions.
2. Proper and Improper Rational Functions
A rational function is classified according to the degrees of its numerator and denominator.
- Proper Rational Function: Degree of the numerator is less than the degree of the denominator.
- Improper Rational Function: Degree of the numerator is greater than or equal to the degree of the denominator.
If the given rational function is improper, first use long division to convert it into a proper rational function before applying partial fractions.
3. Factorisation of the Denominator
Before decomposing a rational function into partial fractions, factorise the denominator completely wherever possible. The form of the denominator determines the type of partial fraction to be used.
- Distinct linear factors
- Repeated linear factors
- Irreducible quadratic factors
4. Partial Fraction Decomposition
The basic idea of this method is to express a complicated rational function as a sum of simpler fractions, each of which can be integrated easily using standard integration formulas.
The required standard forms for partial fraction decomposition are given in the table below for quick reference.
5. Determining the Constants
After decomposing the given rational function, determine the unknown constants A,\;B,\;C,\ldots by comparing coefficients or by substituting suitable values of x. Once these constants are obtained, integrate each simpler fraction separately.
6. Strategy for Solving Rational Function Integrals
- Check whether the rational function is proper or improper.
- If it is improper, first use long division to make it a proper rational function.
- Factorise the denominator completely, wherever possible.
- Choose the appropriate partial fraction form.
- Determine the constants A,\;B,\;C,\ldots.
- Integrate each partial fraction separately and always add the constant of integration C.
Following these steps provides a systematic approach for solving almost every rational function integral in this exercise.
Tip: Before applying partial fractions, first check whether the given rational function is proper or improper. If it is improper, use long division to convert it into a proper rational function. Then factorise the denominator, choose the appropriate partial fraction form, determine the constants, integrate each term separately, and don’t forget to add the constant of integration C.
Partial Fraction Decomposition
Before integrating a rational function, we first express it in the appropriate partial fraction form. The following table shows the most common forms of rational functions and their corresponding partial fraction decompositions used in this exercise.
| Standard Forms for Partial Fraction Decomposition |
|---|
| Form 1: Constant or Linear polynomial divided by two different linear factors or \displaystyle \frac{px+q}{(x-a)(x-b)},\; a\neq b |
| Decompose the function as: \displaystyle \frac{A}{x-a}+\frac{B}{x-b} |
| Form 2: Constant or Linear polynomial divided by two same linear factors. or \displaystyle \frac{px+q}{(x-a)^2} |
| Decompose the function as: \displaystyle \frac{A}{x-a}+\frac{B}{(x-a)^2} |
| Form 3: Constant, Linear or Quadratic polynomial divided by three different linear factors. or \displaystyle \frac{px^2+qx+r}{(x-a)(x-b)(x-c)},\; a,b,c\text{ are distinct} |
| Decompose the function as: \displaystyle \frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c} |
| Form 4: Constant, Linear or Quadratic polynomial divided by three linear factors, two of which are the same. or \displaystyle \frac{px^2+qx+r}{(x-a)^2(x-b)} |
| Decompose the function as: \displaystyle \frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b} |
| Form 5: Constant, Linear or Quadratic polynomial divided by one linear factor and one irreducible quadratic factor. or \displaystyle \frac{px^2+qx+r}{(x-a)(x^2+bx+c)} where x^2+bx+c cannot be factorised further. |
| Decompose the function as: \displaystyle \frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c} |
Important: The above forms are applicable only when the rational function is a proper rational function, i.e., the degree of the numerator is less than the degree of the denominator. If the degree of P(x) is greater than or equal to the degree of Q(x), first use long division to convert it into a proper rational function. Then factorise the denominator (where possible), decompose it into partial fractions, determine the constants A,\;B,\;C,\ldots, and finally integrate each term separately.
Let us now solve all the NCERT questions step by step in this exercise of Integrals 7.5.
Question 1: Integrals 7.5
1. Evaluate \displaystyle \int \frac{x}{(x+1)(x+2)}\,dx.
Solution
Let, \displaystyle I=\int\frac{x}{(x+1)(x+2)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of two different linear factors. Therefore, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x+1)(x+2), we get
\displaystyle x=A(x+2)+B(x+1)
Putting x=-1, we get
\displaystyle -1=A\;\Rightarrow\;A=-1
Putting x=-2, we get
\displaystyle -2=-B\;\Rightarrow\;B=2
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(-\frac{1}{x+1}+\frac{2}{x+2}\right)\,dx
Using the linearity property of integration, we get
\displaystyle I=-\int\frac{dx}{x+1}+2\int\frac{dx}{x+2}
Using \displaystyle \int\frac{dx}{x+a}=\log|x+a|+C, we obtain
\displaystyle I=-\log|x+1|+2\log|x+2|+C
Hence,
\displaystyle I=\int\frac{x}{(x+1)(x+2)}\,dx=-\log|x+1|+2\log|x+2|+C
Using the property \displaystyle n\log m=\log(m^n) and \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\log\left(\frac{(x+2)^2}{|x+1|}\right)+C
Hence,
\boxed{\displaystyle \int\frac{x}{(x+1)(x+2)}\,dx=\log\left(\frac{(x+2)^2}{|x+1|}\right)+C}
Question 2: Integrals Ex. 7.5
2. Evaluate \displaystyle \int\frac{dx}{x^2-9}.
Solution
Let, \displaystyle I=\int\frac{dx}{x^2-9}
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator can be factorised as (x-3)(x+3), which consists of two different linear factors. Therefore, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac{1}{x^2-9}=\frac{A}{x-3}+\frac{B}{x+3}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-3)(x+3), we get
\displaystyle 1=A(x+3)+B(x-3)
Putting x=3, we get
\displaystyle 1=6A\;\Rightarrow\;A=\frac16
Putting x=-3, we get
\displaystyle 1=-6B\;\Rightarrow\;B=-\frac16
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac{1}{6(x-3)}-\frac{1}{6(x+3)}\right)dx
Using the linearity property of integration, we get
\displaystyle I=\frac16\int\frac{dx}{x-3}-\frac16\int\frac{dx}{x+3}
Using \displaystyle \int\frac{dx}{x+a}=\log|x+a|+C, we obtain
\displaystyle I=\frac16\log|x-3|-\frac16\log|x+3|+C
Using the property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\frac16\log\left|\frac{x-3}{x+3}\right|+C
Hence,
\boxed{\displaystyle I=\int\frac{dx}{x^2-9}=\frac16\log\left|\frac{x-3}{x+3}\right|+C}
Alternative Method (Using Standard Integral Formula)
Using the standard result
\displaystyle \int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C
where a=3, we get
\boxed{\displaystyle \int\frac{dx}{x^2-9}=\frac16\log\left|\frac{x-3}{x+3}\right|+C}
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Question 3: Integration by Partial Fractions
3. Evaluate \displaystyle \int\frac{3x-1}{(x-1)(x-2)(x-3)}\,dx.
Solution
Let, \displaystyle I=\int\frac{3x-1}{(x-1)(x-2)(x-3)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of three different linear factors. Therefore, using Form 3 of Partial Fraction Decomposition, we get
\displaystyle \frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-1)(x-2)(x-3), we get
\displaystyle 3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)
Putting x=1, we get
\displaystyle 2=2A\;\Rightarrow\;A=1
Putting x=2, we get
\displaystyle 5=-B\;\Rightarrow\;B=-5
Putting x=3, we get
\displaystyle 8=2C\;\Rightarrow\;C=4
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac{1}{x-1}-\frac{5}{x-2}+\frac{4}{x-3}\right)dx
Using the linearity property of integration, we get
\displaystyle I=\int\frac{dx}{x-1}-5\int\frac{dx}{x-2}+4\int\frac{dx}{x-3}
Using \displaystyle \int\frac{dx}{x+a}=\log|x+a|+C, we obtain
\displaystyle I=\log|x-1|-5\log|x-2|+4\log|x-3|+C
Hence,
\boxed{\begin{gathered} \int\frac{3x-1}{(x-1)(x-2)(x-3)}\,dx\\[8pt] =\log|x-1|-5\log|x-2|+4\log|x-3|+C \end{gathered}}
Question 4: Integrals 7.5
4. Evaluate \displaystyle \int\frac{x}{(x-1)(x-2)(x-3)}\,dx.
Solution
Let, \displaystyle I=\int\frac{x}{(x-1)(x-2)(x-3)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of three different linear factors. Therefore, using Form 3 of Partial Fraction Decomposition, we get
\displaystyle \frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-1)(x-2)(x-3), we get
\displaystyle x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)
Putting x=1, we get
\displaystyle 1=2A\;\Rightarrow\;A=\frac12
Putting x=2, we get
\displaystyle 2=-B\;\Rightarrow\;B=-2
Putting x=3, we get
\displaystyle 3=2C\;\Rightarrow\;C=\frac32
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac{1}{2(x-1)}-\frac{2}{x-2}+\frac{3}{2(x-3)}\right)dx
Using the linearity property of integration, we get
\displaystyle I=\frac12\int\frac{dx}{x-1}-2\int\frac{dx}{x-2}+\frac32\int\frac{dx}{x-3}
Using \displaystyle \int\frac{dx}{x+a}=\log|x+a|+C, we obtain
\displaystyle I=\frac12\log|x-1|-2\log|x-2|+\frac32\log|x-3|+C
Hence,
\boxed{\begin{gathered} \int\frac{x}{(x-1)(x-2)(x-3)}\,dx\\[8pt] =\frac12\log|x-1|-2\log|x-2|+\frac32\log|x-3|+C \end{gathered}}
Question 5: Integration of Proper Rational Function
5. Evaluate \displaystyle \int\frac{2x}{x^2+3x+2}\,dx.
Solution
Let, \displaystyle I=\int\frac{2x}{x^2+3x+2}\,dx
Factorising the denominator, we get
\displaystyle x^2+3x+2=(x+1)(x+2)
Therefore,
\displaystyle I=\int\frac{2x}{(x+1)(x+2)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of two different linear factors. Therefore, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac{2x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x+1)(x+2), we get
\displaystyle 2x=A(x+2)+B(x+1)
Putting x=-1, we get
\displaystyle -2=A\;\Rightarrow\;A=-2
Putting x=-2, we get
\displaystyle -4=-B\;\Rightarrow\;B=4
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(-\frac2{x+1}+\frac4{x+2}\right)\,dx
Using the linearity property of integration, we get
\displaystyle I=-2\int\frac{dx}{x+1}+4\int\frac{dx}{x+2}
Using \displaystyle \int\frac{dx}{x+a}=\log|x+a|+C, we obtain
\displaystyle I=-2\log|x+1|+4\log|x+2|+C
Hence,
\boxed{\displaystyle I=-2\log|x+1|+4\log|x+2|+C}
Tip: Before applying partial fractions, always check whether the given rational function is proper or improper. If it is improper, first convert it into a proper rational function by using long division (or factorisation, whenever convenient), as in Q. 6 of Integrals 7.5 below.
Question 6: Long Division Method
6. Evaluate \displaystyle \int\frac{1-x^2}{x(1-2x)}\,dx.
Solution
Let, \displaystyle I=\int\frac{1-x^2}{x(1-2x)}\,dx
Since the degree of the numerator is equal to the degree of the denominator, the given rational function is an improper rational function. Also, the numerator and denominator have no common factors. Therefore, we first convert it into a proper rational function using long division.
For convenience, we first rewrite the given rational function with a positive leading coefficient in the denominator. Also, before applying long division, the terms of both the numerator and the denominator should be arranged in decreasing powers of x. Thus, we get
\displaystyle \frac{1-x^2}{x(1-2x)}=\frac{x^2-1}{x(2x-1)}=\frac{x^2-1}{2x^2-x}
Using long division as below,
We get the quotient as \displaystyle \frac12 and the remainder as \displaystyle \frac{x}{2}-1. Therefore,
\displaystyle \frac{x^2-1}{2x^2-x}=\frac12+\frac{\frac{x}{2}-1}{2x^2-x}=\frac12+\frac{\frac{x}{2}-1}{x(2x-1)}
Hence,
\displaystyle I=\int\left(\frac12+\frac{\frac12x-1}{x(2x-1)}\right)dx
Since the remaining rational function is proper and the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac{\frac12x-1}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}\qquad\cdots(1)
Multiplying both sides of equation (1) by x(2x-1), we get
\displaystyle \frac12x-1=A(2x-1)+Bx
Putting x=0, we get
\displaystyle -1=-A\;\Rightarrow\;A=1
Comparing the coefficients of x on both sides, we get
\displaystyle \frac12=2A+B=2+B\;\Rightarrow\;B=-\frac32
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac12+\frac1x-\frac{3}{2(2x-1)}\right)dx
Using the linearity property of integration, we get
\displaystyle I=\frac12\int dx+\int\frac{dx}{x}-\frac32\int\frac{dx}{2x-1}
Using \displaystyle \int\frac{dx}{ax+b}=\frac1a\log|ax+b|+C, we obtain
\displaystyle I=\frac{x}{2}+\log|x|-\frac34\log|2x-1|+C
Since |2x-1|=|1-2x|, we may write
\displaystyle I=\frac{x}{2}+\log|x|-\frac34\log|1-2x|+C
Hence,
\boxed{\displaystyle I=\frac{x}{2}+\log|x|-\frac34\log|1-2x|+C}
Question 7: NCERT Exercise Integrals 7.5
7. Evaluate \displaystyle \int\frac{x}{(x^2+1)(x-1)}\,dx.
Solution
Let, \displaystyle I=\int\frac{x}{(x^2+1)(x-1)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of one linear factor and one irreducible quadratic factor. Therefore, using Form 5 of Partial Fraction Decomposition, we get
\displaystyle \frac{x}{(x^2+1)(x-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x^2+1)(x-1), we get
\displaystyle x=A(x^2+1)+(Bx+C)(x-1)
Expanding the right-hand side, we get
\displaystyle x=(A+B)x^2+(-B+C)x+(A-C)
Comparing the coefficients of corresponding powers of x on both sides, we get
\displaystyle A+B=0,\qquad -B+C=1,\qquad A-C=0
Solving these equations, we get
\displaystyle A=\frac12,\qquad B=-\frac12,\qquad C=\frac12
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac{1}{2(x-1)}+\frac{-x+1}{2(x^2+1)}\right)dx
Using the linearity property of integration and splitting the numerator, we get
\displaystyle I=\frac12\int\frac{dx}{x-1}-\frac12\int\frac{x\,dx}{x^2+1}+\frac12\int\frac{dx}{x^2+1}
Multiply and divide second integral by 2, we get
\displaystyle I=\frac12\int\frac{dx}{x-1}-\frac14\int\frac{2x\,dx}{x^2+1}+\frac12\int\frac{dx}{x^2+1}
Using the standard results \displaystyle \int\frac{f'(x)}{f(x)}dx=\log|f(x)|+C and \displaystyle \int\frac{dx}{1+x^2}=\tan^{-1}x+C, we obtain
\displaystyle I=\frac12\log|x-1|-\frac14\log(x^2+1)+\frac12\tan^{-1}x+C
Hence,
\boxed{\begin{gathered} \int\frac{x}{(x^2+1)(x-1)}\,dx\\[8pt] =\frac12\log|x-1|-\frac14\log(x^2+1)+\frac12\tan^{-1}x+C \end{gathered}}
Question 8: Partial Fraction Decomposition
8. Evaluate \displaystyle \int\frac{x}{(x-1)^2(x+2)}\,dx.
Solution
Let, \displaystyle I=\int\frac{x}{(x-1)^2(x+2)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of three linear factors, two of which are identical. Therefore, using Form 4 of Partial Fraction Decomposition, we get
\displaystyle \frac{x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-1)^2(x+2), we get
\displaystyle x=A(x-1)(x+2)+B(x+2)+C(x-1)^2
Putting x=1, we get
\displaystyle 1=3B\;\Rightarrow\;B=\frac13
Putting x=-2, we get
\displaystyle -2=9C\;\Rightarrow\;C=-\frac29
Substituting x=0, we get
\displaystyle 0=-2A+2B+C
Using B=\frac13 and C=-\frac29, we get
\displaystyle A=\frac29
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac{2}{9(x-1)}+\frac{1}{3(x-1)^2}-\frac{2}{9(x+2)}\right)dx
Using the linearity property of integration, we get
\displaystyle I=\frac29\int\frac{dx}{x-1}+\frac13\int\frac{dx}{(x-1)^2}-\frac29\int\frac{dx}{x+2}
Using the standard results \displaystyle \int\frac{dx}{x+a}=\log|x+a|+C and \displaystyle \int\frac{dx}{(x-a)^2}=-\frac1{x-a}+C, we obtain
\displaystyle I=\frac29\log|x-1|-\frac{1}{3(x-1)}-\frac29\log|x+2|+C
Using the property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\frac29\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C
Hence,
\boxed{\displaystyle I=\frac29\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C}
Congratulations on completing NCERT Class 12 Maths Part 1! We have covered all the chapters and exercises with detailed explanations and step-by-step solutions. The journey continues with Part 2, where I’ll keep providing easy-to-follow solutions and concept-based explanations. Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, to help you learn and revise more effectively.
Now, let’s proceed to the next question of Integrals 7.5.
Question 9: Integrals 7.5
9. Evaluate \displaystyle \int\frac{3x+5}{x^3-x^2-x+1}\,dx.
Solution
Let, \displaystyle I=\int\frac{3x+5}{x^3-x^2-x+1}\,dx
Factorising the denominator, we get
\displaystyle x^3-x^2-x+1=x^2(x-1)-(x-1)=(x-1)(x^2-1)=(x-1)^2(x+1)
Therefore,
\displaystyle I=\int\frac{3x+5}{(x-1)^2(x+1)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of three linear factors, two of which are identical. Therefore, using Form 4 of Partial Fraction Decomposition, we get
\displaystyle \frac{3x+5}{(x-1)^2(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-1)^2(x+1), we get
\displaystyle 3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^2
Putting x=1, we get
\displaystyle 8=2B\;\Rightarrow\;B=4
Putting x=-1, we get
\displaystyle 2=4C\;\Rightarrow\;C=\frac12
Substituting x=0, we get
\displaystyle 5=-A+B+C
Using B=4 and C=\frac12, we get
\displaystyle A=-\frac12
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(-\frac{1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}\right)dx
Using the linearity property of integration, we get
\displaystyle I=-\frac12\int\frac{dx}{x-1}+4\int\frac{dx}{(x-1)^2}+\frac12\int\frac{dx}{x+1}
Using the standard results \displaystyle \int\frac{dx}{x+a}=\log|x+a|+C and \displaystyle \int\frac{dx}{(x-a)^2}=-\frac1{x-a}+C, we obtain
\displaystyle I=-\frac12\log|x-1|-\frac4{x-1}+\frac12\log|x+1|+C
Using the property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\frac12\log\left|\frac{x+1}{x-1}\right|-\frac4{x-1}+C
Hence,
\boxed{\displaystyle I=\frac12\log\left|\frac{x+1}{x-1}\right|-\frac4{x-1}+C}
Question 10: Integrals 7.5
10. Evaluate \displaystyle \int\frac{2x-3}{(x^2-1)(2x+3)}\,dx.
Solution
Let, \displaystyle I=\int\frac{2x-3}{(x^2-1)(2x+3)}\,dx
Factorising the denominator, we get
\displaystyle x^2-1=(x-1)(x+1)
Therefore,
\displaystyle I=\int\frac{2x-3}{(x-1)(x+1)(2x+3)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of three different linear factors. Therefore, using Form 3 of Partial Fraction Decomposition, we get
\displaystyle \frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-1)(x+1)(2x+3), we get
\displaystyle 2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)
Putting x=1, we get
\displaystyle -1=10A\;\Rightarrow\;A=-\frac1{10}
Putting x=-1, we get
\displaystyle -5=-2B\;\Rightarrow\;B=\frac52
Substituting x=0, we get
\displaystyle -3=3A-3B-C
Using A=-\frac1{10} and B=\frac52, we get
\displaystyle C=-\frac{24}{5}
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(-\frac1{10(x-1)}+\frac5{2(x+1)}-\frac{24}{5(2x+3)}\right)dx
Using the linearity property of integration, we get
\displaystyle I=-\frac1{10}\int\frac{dx}{x-1}+\frac52\int\frac{dx}{x+1}-\frac{24}{5}\int\frac{dx}{2x+3}
Using \displaystyle \int\frac{dx}{ax+b}=\frac1a\log|ax+b|+C, we obtain
\displaystyle I=-\frac1{10}\log|x-1|+\frac52\log|x+1|-\frac{12}{5}\log|2x+3|+C
Hence,
\boxed{\begin{gathered} \int\frac{2x-3}{(x^2-1)(2x+3)}\,dx\\[6pt] =-\frac1{10}\log|x-1|+\frac52\log|x+1|\\[6pt] -\frac{12}{5}\log|2x+3|+C \end{gathered}}
Tip: Before choosing a partial fraction form, always check whether any quadratic factor can be factorised further. In Q. 10 above, x^2-1=(x-1)(x+1), so we use Form 3, not Form 5. Only those quadratic factors which cannot be factorised further require a numerator of the form Bx+C.
Question 11: Integrals 7.5
11. Evaluate \displaystyle \int\frac{5x}{(x+1)(x^2-4)}\,dx.
Solution
Let, \displaystyle I=\int\frac{5x}{(x+1)(x^2-4)}\,dx
Factorising the denominator, we get
\displaystyle x^2-4=(x-2)(x+2)
Therefore,
\displaystyle I=\int\frac{5x}{(x+1)(x-2)(x+2)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of three different linear factors. Therefore, using Form 3 of Partial Fraction Decomposition, we get
\displaystyle \frac{5x}{(x+1)(x-2)(x+2)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x+1)(x-2)(x+2), we get
\displaystyle 5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)
Putting x=-1, we get
\displaystyle -5=-3A\;\Rightarrow\;A=\frac53
Putting x=2, we get
\displaystyle 10=12B\;\Rightarrow\;B=\frac56
Putting x=-2, we get
\displaystyle -10=4C\;\Rightarrow\;C=-\frac52
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac{5}{3(x+1)}+\frac{5}{6(x-2)}-\frac{5}{2(x+2)}\right)dx
Using the linearity property of integration, we get
\displaystyle I=\frac53\int\frac{dx}{x+1}+\frac56\int\frac{dx}{x-2}-\frac52\int\frac{dx}{x+2}
Using \displaystyle \int\frac{dx}{x+a}=\log|x+a|+C, we obtain
\displaystyle I=\frac53\log|x+1|+\frac56\log|x-2|-\frac52\log|x+2|+C
Hence,
\boxed{\begin{gathered} \int\frac{5x}{(x+1)(x^2-4)}\,dx\\[6pt] =\frac53\log|x+1|+\frac56\log|x-2|\\[6pt] -\frac52\log|x+2|+C \end{gathered}}
Question 12: Improper Rational Function
12. Evaluate \displaystyle \int\frac{x^3+x+1}{x^2-1}\,dx.
Solution
Let, \displaystyle I=\int\frac{x^3+x+1}{x^2-1}\,dx
Since the degree of the numerator is greater than the degree of the denominator, the given rational function is an improper rational function. Also, the numerator and denominator have no common factors. Therefore, we first convert it into a proper rational function using long division.
For convenience, before applying long division, the terms of both the numerator and the denominator are arranged in decreasing powers of x. Thus, we get
\displaystyle \frac{x^3+x+1}{x^2-1}
Using long division as below,
We get the quotient as \displaystyle x and the remainder as \displaystyle 2x+1. Therefore,
\displaystyle \frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}
Factorising the denominator, we get
\displaystyle x^2-1=(x-1)(x+1)
Hence,
\displaystyle I=\int\left(x+\frac{2x+1}{(x-1)(x+1)}\right)dx
Since the remaining rational function is proper and the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac{2x+1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-1)(x+1), we get
\displaystyle 2x+1=A(x+1)+B(x-1)
Putting x=1, we get
\displaystyle 3=2A\;\Rightarrow\;A=\frac32
Putting x=-1, we get
\displaystyle -1=-2B\;\Rightarrow\;B=\frac12
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(x+\frac{3}{2(x-1)}+\frac{1}{2(x+1)}\right)dx
Using the linearity property of integration, we get
\displaystyle I=\int x\,dx+\frac32\int\frac{dx}{x-1}+\frac12\int\frac{dx}{x+1}
Using the standard integration formulas, we obtain
\displaystyle I=\frac{x^2}{2}+\frac32\log|x-1|+\frac12\log|x+1|+C
Hence,
\boxed{\displaystyle I=\frac{x^2}{2}+\frac32\log|x-1|+\frac12\log|x+1|+C}
Question 13: Integrals 7.5
13. Evaluate \displaystyle \int\frac{2}{(1-x)(1+x^2)}\,dx.
Solution
Let, \displaystyle I=\int\frac{2}{(1-x)(1+x^2)}\,dx
For convenience, we first rewrite the given rational function so that the coefficient of x in the linear factor is positive.
\displaystyle \frac{2}{(1-x)(1+x^2)}=-\frac{2}{(x-1)(1+x^2)}
Therefore,
\displaystyle I=\int\frac{-2}{(x-1)(1+x^2)}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of one linear factor and one irreducible quadratic factor. Therefore, using Form 5 of Partial Fraction Decomposition, we get
\displaystyle \frac{-2}{(x-1)(1+x^2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-1)(x^2+1), we get
\displaystyle -2=A(x^2+1)+(Bx+C)(x-1)
Expanding the right-hand side, we get
\displaystyle -2=(A+B)x^2+(C-B)x+(A-C)
Comparing the coefficients of corresponding powers of x on both sides, we get
\displaystyle A+B=0,\qquad C-B=0,\qquad A-C=-2
Solving these equations, we get
\displaystyle A=-1,\qquad B=1,\qquad C=1
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(-\frac1{x-1}+\frac{x+1}{x^2+1}\right)dx
Using the linearity property of integration and splitting the numerator, we get
\displaystyle I=-\int\frac{dx}{x-1}+\int\frac{x\,dx}{x^2+1}+\int\frac{dx}{x^2+1}
Divide and multiply second integral by 2, we get
\displaystyle I=-\int\frac{dx}{x-1}+\frac12\int\frac{2x\,dx}{x^2+1}+\int\frac{dx}{x^2+1}
Using the standard results \displaystyle \int\frac{f'(x)}{f(x)}dx=\log|f(x)|+C and \displaystyle \int\frac{dx}{1+x^2}=\tan^{-1}x+C, we obtain
\displaystyle I=-\log|x-1|+\frac12\log(x^2+1)+\tan^{-1}x+C
Hence,
\boxed{\begin{gathered} \int\frac{2}{(1-x)(1+x^2)}\,dx\\[8pt] =-\log|x-1|+\frac12\log(x^2+1)+\tan^{-1}x+C \end{gathered}}
Tip: In Integrals 7.5, while decomposing into partial fractions, it is usually convenient to keep the coefficient of x positive in every linear factor. This reduces sign errors and makes integration easier.
Question 14: Integrals Exercise 7.5
14. Evaluate \displaystyle \int\frac{3x-1}{(x+2)^2}\,dx.
Solution
Let, \displaystyle I=\int\frac{3x-1}{(x+2)^2}\,dx
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of two identical linear factors. Therefore, using Form 2 of Partial Fraction Decomposition, we get
\displaystyle \frac{3x-1}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x+2)^2, we get
\displaystyle 3x-1=A(x+2)+B
Putting x=-2, we get
\displaystyle -7=B
Comparing the coefficients of x on both sides, we get
\displaystyle A=3
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac{3}{x+2}-\frac{7}{(x+2)^2}\right)dx
Using the linearity property of integration, we get
\displaystyle I=3\int\frac{dx}{x+2}-7\int\frac{dx}{(x+2)^2}
Using the standard results \displaystyle \int\frac{dx}{x+a}=\log|x+a|+C and \displaystyle \int\frac{dx}{(x+a)^2}=-\frac1{x+a}+C, we obtain
\displaystyle I=3\log|x+2|+\frac{7}{x+2}+C
Hence,
\boxed{\displaystyle \int\frac{3x-1}{(x+2)^2}\,dx=3\log|x+2|+\frac{7}{x+2}+C}
Question 15: Indefinite Integrals 7.5
15. Evaluate \displaystyle \int\frac{dx}{x^4-1}.
Solution
Let, \displaystyle I=\int\frac{dx}{x^4-1}
Factorising the denominator, we get
\displaystyle x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)
Therefore,
\displaystyle I=\int\frac{dx}{(x-1)(x+1)(x^2+1)}
Since the degree of the numerator is less than the degree of the denominator, the given rational function is a proper rational function. Also, the denominator consists of two different linear factors and one irreducible quadratic factor. Therefore, using Form 5 of Partial Fraction Decomposition, we get
\displaystyle \frac{1}{(x-1)(x+1)(x^2+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-1)(x+1)(x^2+1), we get
\displaystyle 1=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x^2-1)
Putting x=1, we get
\displaystyle 1=4A\;\Rightarrow\;A=\frac14
Putting x=-1, we get
\displaystyle 1=-4B\;\Rightarrow\;B=-\frac14
Comparing the coefficients of x^3 on both sides, we get
\displaystyle A+B+C=0
Using A=\frac14 and B=-\frac14, we get
\displaystyle C=0
Comparing the coefficients of x^2 on both sides, we get
\displaystyle A-B+D=0
Using A=\frac14 and B=-\frac14, we get
\displaystyle D=-\frac12
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2(x^2+1)}\right)dx
Using the linearity property of integration, we get
\displaystyle I=\frac14\int\frac{dx}{x-1}-\frac14\int\frac{dx}{x+1}-\frac12\int\frac{dx}{x^2+1}
Using the standard integration formulas, we obtain
\displaystyle I=\frac14\log|x-1|-\frac14\log|x+1|-\frac12\tan^{-1}x+C
Using the property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\frac14\log\left|\frac{x-1}{x+1}\right|-\frac12\tan^{-1}x+C
Hence,
\boxed{\displaystyle \int\frac{dx}{x^4-1}=\frac14\log\left|\frac{x-1}{x+1}\right|-\frac12\tan^{-1}x+C}
Tip: In questions involving expressions such as x^n, first look for a suitable substitution. If the required derivative is not present, try multiplying the numerator and denominator by an appropriate power of x to create it. This converts many seemingly difficult integrals into simple rational functions that can be solved using partial fraction as in Q. 16 below.
Question 16: Integrals 7.5
16. Evaluate \displaystyle \int\frac{dx}{x(x^n+1)}.
Solution
Let, \displaystyle I=\int\frac{dx}{x(x^n+1)}
To simplify the integral, we first multiply the numerator and denominator by x^{\,n-1}. This helps us obtain x^n, making the substitution straightforward.
\displaystyle I=\int\frac{x^{\,n-1}\,dx}{x^n(x^n+1)}
Now, put x^n=t. Then,
\displaystyle dt=nx^{\,n-1}dx\qquad\Rightarrow\qquad x^{\,n-1}dx=\frac{dt}{n}
Substituting these values, we get
\displaystyle I=\frac1n\int\frac{dt}{t(t+1)}
Since the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac1{t(t+1)}=\frac{A}{t}+\frac{B}{t+1}\qquad\cdots(1)
Multiplying both sides of equation (1) by t(t+1), we get
\displaystyle 1=A(t+1)+Bt
Putting t=0, we get
\displaystyle 1=A
Putting t=-1, we get
\displaystyle 1=-B\;\Rightarrow\;B=-1
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\frac1n\int\left(\frac1t-\frac1{t+1}\right)dt
Using the linearity property of integration, we get
\displaystyle I=\frac1n\int\frac{dt}{t}-\frac1n\int\frac{dt}{t+1}
Using \displaystyle \int\frac{dt}{t}=\log|t|+C, we obtain
\displaystyle I=\frac1n\log|t|-\frac1n\log|t+1|+C
Substituting t=x^n and using the property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\frac1n\log\left|\frac{x^n}{x^n+1}\right|+C
Hence,
\boxed{\displaystyle \int\frac{dx}{x(x^n+1)}=\frac1n\log\left|\frac{x^n}{x^n+1}\right|+C}
Question 17: Substitution Method – Integrals 7.5
17. Evaluate \displaystyle \int\frac{\cos x}{(1-\sin x)(2-\sin x)}\,dx.
Solution
Let, \displaystyle I=\int\frac{\cos x}{(1-\sin x)(2-\sin x)}\,dx
To simplify the integral, we use the substitution \sin x=t. Then,
\displaystyle dt=\cos x\,dx
Substituting these values, we get
\displaystyle I=\int\frac{dt}{(1-t)(2-t)}
For convenience, we first rewrite the denominator so that the coefficient of t in each linear factor is positive.
\displaystyle I=\int\frac{dt}{(t-1)(t-2)}
Since the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac1{(t-1)(t-2)}=\frac{A}{t-1}+\frac{B}{t-2}\qquad\cdots(1)
Multiplying both sides of equation (1) by (t-1)(t-2), we get
\displaystyle 1=A(t-2)+B(t-1)
Putting t=1, we get
\displaystyle 1=-A\;\Rightarrow\;A=-1
Putting t=2, we get
\displaystyle 1=B
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(-\frac1{t-1}+\frac1{t-2}\right)dt
Using the linearity property of integration, we get
\displaystyle I=-\int\frac{dt}{t-1}+\int\frac{dt}{t-2}
Using the standard integration formula, we obtain
\displaystyle I=-\log|t-1|+\log|t-2|+C
Substituting t=\sin x and using the property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\log\left|\frac{\sin x-2}{\sin x-1}\right|+C
Since |\sin x-2|=|2-\sin x| and |\sin x-1|=|1-\sin x|, we may write
\displaystyle I=\log\left|\frac{2-\sin x}{1-\sin x}\right|+C
Hence,
\boxed{\displaystyle I=\log\left|\frac{2-\sin x}{1-\sin x}\right|+C}
Tip: Whenever the integrand contains a function and its derivative together, such as \sin x with \cos x\,dx, or \cos x with -\sin x\,dx, first look for a suitable substitution. The resulting integral often becomes a rational function that can be evaluated using partial fractions.
Question 18: Integrals 7.5
18. Evaluate \displaystyle \int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx.
Solution
Let, \displaystyle I=\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx
To simplify the partial fraction decomposition, we first put t=x^2. This substitution is used only for convenience in the algebra. After decomposition, we shall substitute t=x^2 back and continue the integration with respect to x.
\displaystyle \frac{(t+1)(t+2)}{(t+3)(t+4)}=\frac{t^2+3t+2}{t^2+7t+12}\qquad\cdots(1)
Since the degree of the numerator is equal to the degree of the denominator, we first perform the division.
\displaystyle \frac{t^2+3t+2}{t^2+7t+12} = \frac{(t^2+7t+12)-(4t+10)}{t^2+7t+12}
Splitting the numerator on RHS,
\displaystyle = \frac{t^2+7t+12}{t^2+7t+12} +\frac{-4t-10}{t^2+7t+12}
Rewriting the denominator as product of linear factors,
We get equation (1) as \displaystyle \frac{t^2+3t+2}{t^2+7t+12}= 1+\frac{-4t-10}{(t+3)(t+4)}\qquad\cdots(2)
Since the remaining rational function is proper and the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac{-4t-10}{(t+3)(t+4)}=\frac{A}{t+3}+\frac{B}{t+4}\qquad\cdots(3)
Multiplying both sides of equation (3) by (t+3)(t+4), we get
\displaystyle -4t-10=A(t+4)+B(t+3)
Putting t=-3, we get
\displaystyle 2=A\;\Rightarrow\;A=2
Putting t=-4, we get
\displaystyle 6=-B\;\Rightarrow\;B=-6
Substituting these values in equation (3), (2) and (1), we get
\displaystyle \frac{(t+1)(t+2)}{(t+3)(t+4)}=1+\frac2{t+3}-\frac6{t+4}
Putting t=x^2 back, we obtain
\displaystyle \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1+\frac2{x^2+3}-\frac6{x^2+4}
Therefore, integrating both sides, we get
\displaystyle I=\int\left(1+\frac2{x^2+3}-\frac6{x^2+4}\right)dx
Using the linearity property of integration, we get
\displaystyle I=\int dx+2\int\frac{dx}{x^2+3}-6\int\frac{dx}{x^2+4}
Using the standard result \displaystyle \int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\!\left(\frac xa\right)+C, we obtain
\displaystyle I=x+\frac2{\sqrt3}\tan^{-1}\!\left(\frac{x}{\sqrt3}\right)-3\tan^{-1}\!\left(\frac x2\right)+C
Hence,
\boxed{\begin{gathered} \int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx\\[6pt] =x+\frac2{\sqrt3}\tan^{-1}\!\left(\frac{x}{\sqrt3}\right)\\[6pt] -3\tan^{-1}\!\left(\frac x2\right)+C \end{gathered}}
Tip: Sometimes, replacing a repeated expression such as x^2 by a new variable helps simplify the algebra involved in partial fraction decomposition. In Q. 18 above, t=x^2 is introduced only for convenience. After decomposition, we substitute t=x^2 back and continue the integration with respect to x.
Question 19: Integrals Exercise 7.5
19. Evaluate \displaystyle \int\frac{2x}{(x^2+1)(x^2+3)}\,dx.
Solution
Let, \displaystyle I=\int\frac{2x}{(x^2+1)(x^2+3)}\,dx
To simplify the integral, we put x^2=t. Then,
\displaystyle dt=2x\,dx
Substituting these values, we get
\displaystyle I=\int\frac{dt}{(t+1)(t+3)}
Since the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}\qquad\cdots(1)
Multiplying both sides of equation (1) by (t+1)(t+3), we get
\displaystyle 1=A(t+3)+B(t+1)
Putting t=-1, we get
\displaystyle 1=2A\;\Rightarrow\;A=\frac12
Putting t=-3, we get
\displaystyle 1=-2B\;\Rightarrow\;B=-\frac12
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(\frac1{2(t+1)}-\frac1{2(t+3)}\right)dt
Using the linearity property of integration, we get
\displaystyle I=\frac12\int\frac{dt}{t+1}-\frac12\int\frac{dt}{t+3}
Using \displaystyle \int\frac{dt}{t+a}=\log|t+a|+C, we obtain
\displaystyle I=\frac12\log|t+1|-\frac12\log|t+3|+C
Substituting t=x^2 and using the property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\frac12\log\left(\frac{x^2+1}{x^2+3}\right)+C
Hence,
\boxed{\displaystyle \int\frac{2x}{(x^2+1)(x^2+3)}\,dx=\frac12\log\left(\frac{x^2+1}{x^2+3}\right)+C}
Tip: In Q. 19 above, t=x^2 is a substitution, not merely a convenience for algebra as in Q. 18. Since dt=2x\,dx is present in the integral, we first change the variable of integration from x to t, complete the integration entirely in terms of t, and finally substitute t=x^2 back in the answer.
Question 20: Special Integrals
20. Evaluate \displaystyle \int\frac{dx}{x(x^4-1)}
Solution
Let, \displaystyle I=\int\frac{dx}{x(x^4-1)}
To simplify the integral, we first multiply the numerator and denominator by x^3. This helps us obtain x^4, making the substitution straightforward.
\displaystyle I=\int\frac{x^3\,dx}{x^4(x^4-1)}
Now, put x^4=t. Then,
\displaystyle dt=4x^3\,dx\qquad\Rightarrow\qquad x^3\,dx=\frac{dt}{4}
Substituting these values, we get
\displaystyle I=\frac14\int\frac{dt}{t(t-1)}
Since the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac1{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}\qquad\cdots(1)
Multiplying both sides of equation (1) by t(t-1), we get
\displaystyle 1=A(t-1)+Bt
Putting t=0, we get
\displaystyle 1=-A\;\Rightarrow\;A=-1
Putting t=1, we get
\displaystyle 1=B
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\frac14\int\left(-\frac1t+\frac1{t-1}\right)dt
Using the linearity property of integration, we get
\displaystyle I=-\frac14\int\frac{dt}{t}+\frac14\int\frac{dt}{t-1}
Using \displaystyle \int\frac{dt}{t}\,dt=\log|t|+C, we obtain
\displaystyle I=-\frac14\log|t|+\frac14\log|t-1|+C
Substituting t=x^4 and using the property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\frac14\log\left|\frac{x^4-1}{x^4}\right|+C
Hence,
\boxed{\displaystyle \int\frac{dx}{x(x^4-1)}=\frac14\log\left|\frac{x^4-1}{x^4}\right|+C}
Tip: Whenever the denominator contains expressions such as x(x^n\pm1), try multiplying the numerator and denominator by x^{\,n-1}. This creates x^n, making the substitution x^n=t straightforward and reducing the integral to a simple rational function.
Question 21: Exponential Integrals 7.5
21. Evaluate \displaystyle \int\frac{dx}{e^x-1}
Solution
Let, \displaystyle I=\int\frac{dx}{e^x-1}
To simplify the integral, we put e^x=t. Then,
\displaystyle dt=e^x\,dx=t\,dx\qquad\Rightarrow\qquad dx=\frac{dt}{t}
Substituting these values, we get
\displaystyle I=\int\frac{dt}{t(t-1)}
Since the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac1{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}\qquad\cdots(1)
Multiplying both sides of equation (1) by t(t-1), we get
\displaystyle 1=A(t-1)+Bt
Putting t=0, we get
\displaystyle 1=-A\;\Rightarrow\;A=-1
Putting t=1, we get
\displaystyle 1=B
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(-\frac1t+\frac1{t-1}\right)dt
Using the linearity property of integration, we get
\displaystyle I=-\int\frac{dt}{t}+\int\frac{dt}{t-1}
Using \displaystyle \int\frac{dt}{t}=\log|t|+C, we obtain
\displaystyle I=-\log|t|+\log|t-1|+C
Substituting t=e^x and using the property \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\log\left|\frac{e^x-1}{e^x}\right|+C
Hence,
\boxed{\displaystyle \int\frac{dx}{e^x-1}=\log\left|\frac{e^x-1}{e^x}\right|+C}
Tip: Whenever the integrand contains an exponential expression such as e^x, first check whether the substitution e^x=t (or a^x=t) simplifies the integral. The resulting integral often reduces to a rational function that can be evaluated using partial fractions.
Now, let’s solve the two important MCQ’s of Integrals 7.5.
Question 22: Integrals 7.5 – MCQ
22. Choose the correct answer.
\displaystyle \int\frac{x}{(x-1)(x-2)}\,dx equals
- (A) \displaystyle \log\left|\frac{(x-1)^2}{x-2}\right|+C
- (B) \displaystyle \log\left|\frac{(x-2)^2}{x-1}\right|+C
- (C) \displaystyle \log\left|\frac{(x-1)^2}{x-2}\right|+C
- (D) \displaystyle \log|(x-1)(x-2)|+C
Solution
Method 1: Using Partial Fraction Decomposition
Let, \displaystyle I=\int\frac{x}{(x-1)(x-2)}\,dx
Since the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\qquad\cdots(1)
Multiplying both sides of equation (1) by (x-1)(x-2), we get
\displaystyle x=A(x-2)+B(x-1)
Putting x=1, we get
\displaystyle 1=-A\;\Rightarrow\;A=-1
Putting x=2, we get
\displaystyle 2=B
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\int\left(-\frac1{x-1}+\frac2{x-2}\right)dx
\displaystyle I=-\log|x-1|+2\log|x-2|+C
Using the properties of logarithms, we get
\boxed{\displaystyle I=\log\left|\frac{(x-2)^2}{x-1}\right|+C}
✅️ Hence, the correct answer is (B).
Method 2: MathsBetter Shortcut – Rule of “Everywhere, Not There!”
This question can also be solved much faster using an interesting Rule of “Everywhere, Not There!”. It works for proper rational functions having distinct linear factors.
For the partial fraction
\displaystyle \frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}
To find the coefficient of a denominator factor, substitute its root in everywhere except that factor.
Therefore,
\displaystyle A=\frac{x}{x-2}\Bigg|_{x=1}=\frac{1}{1-2}=-1
\displaystyle B=\frac{x}{x-1}\Bigg|_{x=2}=\frac{2}{2-1}=2
Hence,
\displaystyle \frac{x}{(x-1)(x-2)}=-\frac1{x-1}+\frac2{x-2}
Integrating both sides, we get
\displaystyle I=-\log|x-1|+2\log|x-2|+C
\boxed{\displaystyle I=\log\left|\frac{(x-2)^2}{x-1}\right|+C}
✅️ Hence, the correct answer is (B).
MathsBetter Tip: The Rule of “Everywhere, Not There!” is a quick way to find the constants in partial fractions involving distinct linear factors. For a detailed explanation, proof, examples and situations where this rule does not apply, read the article “Partial Fractions for Integration – Rule of Everywhere, Not There!“ and/or watch this video on YouTube.
Question 23: Integrals 7.5 – MCQ
23. Choose the correct answer.
\displaystyle \int\frac{dx}{x(x^2+1)} equals
- (A) \displaystyle \log|x|-\frac12\log(x^2+1)+C
- (B) \displaystyle \log|x|+\frac12\log(x^2+1)+C
- (C) \displaystyle -\log|x|+\frac12\log(x^2+1)+C
- (D) \displaystyle \frac12\log|x|+\log(x^2+1)+C
Solution
Let, \displaystyle I=\int\frac{dx}{x(x^2+1)}
Instead of decomposing the given rational function directly, we first multiply the numerator and denominator by x. This allows us to use the substitution x^2=t, making the partial fraction decomposition much simpler.
\displaystyle I=\int\frac{x\,dx}{x^2(x^2+1)}
Now, put x^2=t. Then,
\displaystyle dt=2x\,dx\qquad\Rightarrow\qquad x\,dx=\frac{dt}{2}
Substituting these values, we get
\displaystyle I=\frac12\int\frac{dt}{t(t+1)}
Since the denominator consists of two different linear factors, using Form 1 of Partial Fraction Decomposition, we get
\displaystyle \frac1{t(t+1)}=\frac{A}{t}+\frac{B}{t+1}\qquad\cdots(1)
Multiplying both sides of equation (1) by t(t+1), we get
\displaystyle 1=A(t+1)+Bt
Putting t=0, we get
\displaystyle 1=A
Putting t=-1, we get
\displaystyle 1=-B\;\Rightarrow\;B=-1
Substituting these values in equation (1) and integrating both sides, we get
\displaystyle I=\frac12\int\left(\frac1t-\frac1{t+1}\right)dt
Using the linearity property of integration, we get
\displaystyle I=\frac12\int\frac{dt}{t}-\frac12\int\frac{dt}{t+1}
Using the standard integration formulas, we obtain
\displaystyle I=\frac12\log|t|-\frac12\log(t+1)+C
Substituting t=x^2 and using \log|x^2|=2\log|x|, we get
\boxed{\displaystyle I=\log|x|-\frac12\log(x^2+1)+C}
✅️ Hence, the correct answer is (A).
MathsBetter Tip: Before applying partial fractions, always look for a way to simplify the integral. Here, multiplying the numerator and denominator by x reduces the problem to a much simpler partial fraction after the substitution x^2=t, avoiding the longer decomposition \displaystyle \frac{A}{x}+\frac{Bx+C}{x^2+1}.
MathsBetter Speed Trick: Whenever the denominator consists of two consecutive linear factors, such as x(x+1), (x+2)(x+3) or (x-5)(x-4), the partial fractions can often be written immediately by inspection.
For example,
\displaystyle \frac{1}{x(x+1)}=\frac1x-\frac1{x+1}
The idea is simple: since the two factors are consecutive, their difference is 1. Hence, the partial fractions are simply the reciprocal of the smaller factor minus the reciprocal of the bigger factor.
This quick observation is especially useful in MCQs and competitive exams, where recognizing such patterns can save valuable time.
Common Mistakes to Avoid
- Not checking whether the rational function is proper or improper: If the degree of the numerator is greater than or equal to the degree of the denominator, first use long division to convert it into a proper rational function before applying partial fractions.
- Using the wrong partial fraction form: Choose the correct form based on the factors of the denominator. Different linear factors, repeated linear factors and irreducible quadratic factors all require different decompositions.
- Forgetting to factorise the denominator completely: Before decomposing, always check whether quadratic or higher-degree expressions can be factorised further. This often leads to a much simpler partial fraction.
- Making mistakes while finding the constants: After multiplying through by the denominator, substitute suitable values or compare coefficients carefully. Verify the values before proceeding to integration.
- Ignoring a simpler substitution: In some questions, multiplying the numerator and denominator by a suitable expression and then using a substitution such as x^2=t or x^n=t can reduce the work considerably.
- Forgetting to simplify the partial fractions before integrating: Once the decomposition is complete, rewrite the integrand in its simplest form before applying the standard integration formulas.
- Making errors in logarithmic integrals: Remember that \displaystyle \int\frac{dx}{ax+b}=\frac1a\log|ax+b|+C. Also, use absolute value signs wherever required.
- Combining logarithms incorrectly: Apply the properties of logarithms carefully. For example, \displaystyle \log m-\log n=\log\left(\frac{m}{n}\right) and \displaystyle k\log m=\log(m^k).
- Forgetting the constant of integration: Every indefinite integral must end with the arbitrary constant C. Omitting it makes the answer incomplete.
- Not checking the final answer: Differentiate your result whenever possible. If differentiation gives back the original integrand, the integration is correct.
Continue Learning
After completing Exercise 7.5 of Integrals, you should now be familiar with the method of integrating rational functions using partial fraction decomposition. You have also learned how to convert improper rational functions into proper ones, choose the correct partial fraction form, and combine substitution with partial fractions whenever required.
To strengthen your understanding further, make sure that you revise:
- How to identify whether a rational function is proper or improper
- Converting an improper rational function into a proper one using long division
- Factorising the denominator completely before applying partial fractions
- Choosing the correct partial fraction form based on distinct linear factors, repeated linear factors and irreducible quadratic factors
- Finding the unknown constants by substitution or by comparing coefficients
- Applying standard integration formulas after completing the partial fraction decomposition
- Using logarithmic properties to simplify the final answer whenever required
- Recognising when a suitable substitution, such as x^2=t or x^n=t, can simplify the integration before applying partial fractions
- Quick techniques such as the Rule of “Everywhere, Not There!” and other useful pattern-recognition tricks for MCQs and competitive exams
- Checking the final answer by differentiating it whenever possible
Explore More
Partial fractions become much easier with practice. As you solve more questions, first identify whether the given rational function is proper or improper, choose the correct partial fraction form, and simplify the expression whenever possible before integrating. Look for opportunities to use suitable substitutions and logarithmic properties to reduce your work. With regular practice, you’ll quickly recognize common patterns and solve even lengthy integration problems with greater speed and confidence.
All the best and keep learning 👍
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