Integrals: Chapter 7 Links
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Integrals 7.4 introduces some important special integral formulas that are frequently used in calculus. Many integrals involving quadratic expressions can be transformed into these standard forms and evaluated directly.
\displaystyle \int \frac{dx}{x^2+a^2},\quad \int \frac{dx}{a^2-x^2},\quad \int \frac{dx}{\sqrt{a^2-x^2}}
In this exercise, we will learn how to identify these standard forms and apply the corresponding formulas. Most questions can be solved by first reducing the given expression to one of the six special integrals using a suitable substitution or by completing the square.
The integrals in this exercise generally involve expressions of the form constant or linear expression divided by a quadratic expression or constant or linear expression divided by the square root of a quadratic expression. Once the required form is obtained, the answer follows directly from the appropriate standard result.
Key Concepts
1. Special Integrals
This exercise introduces six important standard integral formulas. Many complicated-looking integrals can be reduced to one of these forms and evaluated directly.
Whenever the integrand matches a standard form, we can apply the corresponding formula instead of integrating from first principles.
2. Recognising Standard Forms
The most important skill in this exercise is identifying whether a given integral can be transformed into one of the standard forms.
- Constant or linear expression divided by a quadratic expression.
- Constant or linear expression divided by the square root of a quadratic expression.
- Expressions that become standard forms after a simple substitution.
3. Linear Substitution
Many integrals can be simplified by replacing a linear expression with a new variable.
For example, if the integrand contains ax+b, we may put t=ax+b. Then dt=a\,dx, which converts the integral into a simpler standard form.
4. Completing the Square
When a quadratic expression does not immediately match a standard formula, we often rewrite it by completing the square.
For example,
x^2+6x+13=(x+3)^2+4
This helps convert the integral into one of the standard forms involving x^2+a^2, a^2-x^2 or x^2-a^2.
5. Standard Results Used in This Exercise
Before solving the questions, learn the six special integral formulas given in the NCERT textbook. Most answers in this exercise are obtained by reducing the given integral to one of these standard results.
6. Strategy for Solving Questions
- Identify the standard form hidden in the integrand.
- If required, use a linear substitution.
- Complete the square when dealing with quadratic expressions.
- Match the resulting integral with the appropriate special integral formula.
- Apply the formula and add the constant of integration C.
Special Integral Formulas
The following standard results are used throughout Exercise 7.4. Learn these formulas carefully and try to identify which standard form matches the given integral.
| S.No. | Special Integral Formula |
|---|---|
| 1. | \displaystyle \int \frac{dx}{x^2-a^2}=\frac1{2a}\log\left|\frac{x-a}{x+a}\right|+C |
| T.P. Put x=a\sec\theta (or x=a\cosec\theta) or use Partial Fractions (Ex. 7.5). | |
| 2. | \displaystyle \int \frac{dx}{a^2-x^2}=\frac1{2a}\log\left|\frac{a+x}{a-x}\right|+C |
| T.P. Put x=a\sin\theta (or x=a\cos\theta) or use Partial Fractions (Ex. 7.5). | |
| 3. | \displaystyle \int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C |
| T.P. Put x=a\tan\theta (or x=a\cot\theta) | |
| 4. | \displaystyle \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C |
| T.P. Put x=a\sin\theta (or x=a\cos\theta). | |
| 5. | \displaystyle \int \frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C |
| T.P. Put x=a\sec\theta (or x=a\cosec\theta) | |
| 6. | \displaystyle \int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C |
| T.P. Put x=a\tan\theta (or x=a\cot\theta) |
Common Integral Forms
Most questions in this exercise belong to one of the following categories. The goal is to transform the given integral into a standard form and then apply the appropriate result.
| Type of Integral | |
|---|---|
| 1 | \displaystyle \int \frac{\text{Constant}}{\text{Quadratic}}\,dx e.g. \displaystyle \int \frac{dx}{ax^2+bx+c} |
| 2 | \displaystyle \int \frac{\text{Constant}}{\sqrt{\text{Quadratic}}}\,dx e.g. \displaystyle \int \frac{dx}{\sqrt{ax^2+bx+c}} |
| Method: Complete the square in the denominator and reduce the integral to one of the six standard forms listed above. | |
| 3 | \displaystyle \int \frac{\text{Linear}}{\text{Quadratic}}\,dx e.g. \displaystyle \int \frac{px+q}{ax^2+bx+c}\,dx |
| 4 | \displaystyle \int \frac{\text{Linear}}{\sqrt{\text{Quadratic}}}\,dx e.g. \displaystyle \int \frac{px+q}{\sqrt{ax^2+bx+c}}\,dx |
|
Method: Find constants A and B such that
\text{Linear}=A(\text{Derivative of Denominator})+B e.g. px+q=A(2ax+b)+B Split the integral into two parts, say I_1 and I_2. Then I_1 usually becomes an \frac{f'(x)}{f(x)} type integral, while I_2 reduces to Type 1 or Type 2 above. |
Tip: In Exercise 7.4, the hardest part is usually not integration. The real challenge is recognizing the correct standard form. Whenever a quadratic expression appears in the denominator or inside a square root, first try completing the square. This often converts the integral into one of the six special integrals listed above.
Let us now solve all the NCERT questions step by step in this exercise 7.4 of Integrals.
Question 1: Integrals 7.4
1. Evaluate \displaystyle \int \frac{3x^2}{x^6+1}\,dx.
Solution
Let, \displaystyle I=\int \frac{3x^2}{x^6+1}\,dx
Rewrite the denominator as
\displaystyle x^6+1=(x^3)^2+1
Therefore,
\displaystyle I=\int \frac{3x^2}{(x^3)^2+1}\,dx
Since the numerator contains 3x^2, which is the derivative of x^3, let
t=x^3
Then
\displaystyle dt=3x^2\,dx
Substituting, we get
\displaystyle I=\int \frac{dt}{t^2+1}
This is of the type
\displaystyle \int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with a=1, we get
\displaystyle I=\tan^{-1}t+C
Substituting t=x^3,
\displaystyle I=\tan^{-1}(x^3)+C
Hence,
\boxed{\displaystyle I=\tan^{-1}(x^3)+C}
Question 2: Integrals Ex. 7.4
2. Evaluate \displaystyle \int \frac{dx}{\sqrt{1+4x^2}}.
Solution
Let, \displaystyle I=\int \frac{dx}{\sqrt{1+4x^2}}
Rewrite the expression inside the square root as
\displaystyle 1+4x^2=(2x)^2+1
Therefore,
\displaystyle I=\int \frac{dx}{\sqrt{(2x)^2+1}}
Let
t=2x
Then
\displaystyle dt=2\,dx or \displaystyle dx=\frac{dt}{2}
Substituting, we get
\displaystyle I=\frac12\int \frac{dt}{\sqrt{t^2+1}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C
Using the above formula with a=1, we get
\displaystyle I=\frac12\log\left|t+\sqrt{t^2+1}\right|+C
Substituting t=2x,
\displaystyle I=\frac12\log\left|2x+\sqrt{1+4x^2}\right|+C
Hence,
\boxed{\displaystyle \int \frac{dx}{\sqrt{1+4x^2}}=\frac12\log\left|2x+\sqrt{1+4x^2}\right|+C}
You may already be following Maths Better for NCERT Solutions for the topics like
- Matrices
- Determinants
- Relations and Functions
- Inverse Trigonometric Functions
- Continuity and Differentiability
- Application of Derivatives
Likewise this exercise of Integrals 7.4 for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills. Now, let’s proceed to the next question.
Question 3: Completing the Square
3. Evaluate \displaystyle \int \frac{dx}{\sqrt{(2-x)^2+1}}.
Solution
Let, \displaystyle I=\int \frac{dx}{\sqrt{(2-x)^2+1}}
Since the expression inside the square root contains (2-x), let
t=2-x
Then
\displaystyle dt=-dx or \displaystyle dx=-dt
Substituting, we get
\displaystyle I=-\int \frac{dt}{\sqrt{t^2+1}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C
Using the above formula with a=1, we get
\displaystyle I=-\log\left|t+\sqrt{t^2+1}\right|+C
Substituting t=2-x,
\displaystyle I=-\log\left|2-x+\sqrt{(2-x)^2+1}\right|+C
Thus,
\displaystyle \int \frac{dx}{\sqrt{(2-x)^2+1}}=-\log\left|2-x+\sqrt{(2-x)^2+1}\right|+C
Using the identity
\displaystyle -\log|A|=\log\left|\frac1A\right|
we get
\displaystyle I=\log\left|\frac1{2-x+\sqrt{(2-x)^2+1}}\right|+C
or \displaystyle I=\log\left|\frac1{2-x+\sqrt{x^2-4x+5}}\right|+C
Hence,
\boxed{\displaystyle I=\log\left|\frac1{2-x+\sqrt{x^2-4x+5}}\right|+C}
Question 4: Integrals 7.4
4. Evaluate \displaystyle \int \frac{dx}{\sqrt{9-25x^2}}.
Solution
Let, \displaystyle I=\int \frac{dx}{\sqrt{9-25x^2}}
Rewrite the expression inside the square root as
\displaystyle 9-25x^2=3^2-(5x)^2
Therefore,
\displaystyle I=\int \frac{dx}{\sqrt{3^2-(5x)^2}}
Let t=5x
Then \displaystyle dt=5\,dx or \displaystyle dx=\frac{dt}{5}
Substituting, we get
\displaystyle I=\frac15\int \frac{dt}{\sqrt{3^2-t^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with a=3, we get
\displaystyle I=\frac15\sin^{-1}\left(\frac{t}{3}\right)+C
Substituting t=5x,
\displaystyle I=\frac15\sin^{-1}\left(\frac{5x}{3}\right)+C
Hence,
\boxed{\displaystyle \int \frac{dx}{\sqrt{9-25x^2}}=\frac15\sin^{-1}\left(\frac{5x}{3}\right)+C}
Question 5: Method of Substitution
5. Evaluate \displaystyle \int \frac{3x}{1+2x^4}\,dx.
Solution
Let, \displaystyle I=\int \frac{3x}{1+2x^4}\,dx
Rewrite the denominator as
\displaystyle 1+2x^4=1+(\sqrt2\,x^2)^2
Therefore,
\displaystyle I=\int \frac{3x}{1+(\sqrt2\,x^2)^2}\,dx
Let t=\sqrt2\,x^2
Then \displaystyle dt=2\sqrt2\,x\,dx or \displaystyle x\,dx=\frac{dt}{2\sqrt2}
Substituting, we get
\displaystyle I=\frac{3}{2\sqrt2}\int \frac{dt}{1+t^2}
This is of the type
\displaystyle \int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with a=1, we get
\displaystyle I=\frac{3}{2\sqrt2}\tan^{-1}t+C
Substituting t=\sqrt2\,x^2,
\displaystyle I=\frac{3}{2\sqrt2}\tan^{-1}(\sqrt2\,x^2)+C
Hence,
\boxed{\displaystyle \int \frac{3x}{1+2x^4}\,dx=\frac{3}{2\sqrt2}\tan^{-1}(\sqrt2\,x^2)+C}
Important: Whenever a quadratic expression of the form ax^2+bx+c appears in the denominator or under a square root, your first goal should be to complete the square. This often reduces the integral to one of the standard forms involving a^2-x^2, a^2+x^2 or x^2-a^2, making the integration much easier.
Question 6: Integrals 7.4
6. Evaluate \displaystyle \int \frac{x^2}{1-x^6}\,dx.
Solution
Let, \displaystyle I=\int \frac{x^2}{1-x^6}\,dx
Rewrite the denominator as
\displaystyle 1-x^6=1-(x^3)^2
Therefore,
\displaystyle I=\int \frac{x^2}{1-(x^3)^2}\,dx
Since the numerator contains x^2, which is proportional to the derivative of x^3,
Let t=x^3
Then \displaystyle dt=3x^2\,dx or \displaystyle x^2\,dx=\frac{dt}{3}
Substituting, we get
\displaystyle I=\frac13\int \frac{dt}{1-t^2}
This is of the type
\displaystyle \int \frac{dx}{a^2-x^2}=\frac1{2a}\log\left|\frac{a+x}{a-x}\right|+C
Using the above formula with a=1, we get
\displaystyle I=\frac13\left(\frac12\log\left|\frac{1+t}{1-t}\right|\right)+C
\displaystyle I=\frac16\log\left|\frac{1+t}{1-t}\right|+C
Substituting t=x^3,
\displaystyle I=\frac16\log\left|\frac{1+x^3}{1-x^3}\right|+C
Hence,
\boxed{\displaystyle \int \frac{x^2}{1-x^6}\,dx=\frac16\log\left|\frac{1+x^3}{1-x^3}\right|+C}
Question 7: Exercise 7.4
7. Evaluate \displaystyle \int \frac{x-1}{\sqrt{x^2-1}}\,dx.
Solution
Let, \displaystyle I=\int \frac{x-1}{\sqrt{x^2-1}}\,dx
Splitting the Numerator, we get
\displaystyle I=\int \frac{x\,dx}{\sqrt{x^2-1}}-\int \frac{dx}{\sqrt{x^2-1}}
Let
\displaystyle I=I_1-I_2 …(1)
where
\displaystyle I_1=\int \frac{x\,dx}{\sqrt{x^2-1}}
and
\displaystyle I_2=\int \frac{dx}{\sqrt{x^2-1}}
To evaluate I_1,
Let t=x^2-1
Then \displaystyle dt=2x\,dx
or \displaystyle x\,dx=\frac{dt}{2}
Therefore,
\displaystyle I_1=\frac12\int \frac{dt}{\sqrt t}=\frac12\int t^{-1/2}\,dt
\displaystyle \Rightarrow I_1=\sqrt t
i.e. \displaystyle I_1=\sqrt{x^2-1} …(2)
To evaluate I_2, note that
\displaystyle I_2=\int \frac{dx}{\sqrt{x^2-1}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C
Using the above formula with a=1, we get
\displaystyle I_2=\log\left|x+\sqrt{x^2-1}\right| …(3)
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=\sqrt{x^2-1}-\log\left|x+\sqrt{x^2-1}\right|+C
Hence,
\boxed{\displaystyle I=\sqrt{x^2-1}-\log\left|x+\sqrt{x^2-1}\right|+C}
Question 8: Standard Form of Integrals
8. Evaluate \displaystyle \int \frac{x^2}{\sqrt{x^6+a^6}}\,dx.
Solution
Let, \displaystyle I=\int \frac{x^2}{\sqrt{x^6+a^6}}\,dx
Rewrite the expression inside the square root as
\displaystyle x^6+a^6=(x^3)^2+(a^3)^2
Therefore,
\displaystyle I=\int \frac{x^2\,dx}{\sqrt{(x^3)^2+(a^3)^2}}
Since the numerator contains x^2, which is proportional to the derivative of x^3,
Let t=x^3
Then \displaystyle dt=3x^2\,dx
or \displaystyle x^2\,dx=\frac{dt}{3}
Substituting, we get
\displaystyle I=\frac13\int \frac{dt}{\sqrt{t^2+(a^3)^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C
Using the above formula with a=a^3, we get
\displaystyle I=\frac13\log\left|t+\sqrt{t^2+a^6}\right|+C
Substituting t=x^3,
\displaystyle I=\frac13\log\left|x^3+\sqrt{x^6+a^6}\right|+C
Hence,
\boxed{\displaystyle \int \frac{x^2}{\sqrt{x^6+a^6}}\,dx=\frac13\log\left|x^3+\sqrt{x^6+a^6}\right|+C}
By now, you have probably noticed that the key to most questions in this exercise is recognizing the pattern hidden inside the integrand. Once the expression is reduced to a standard form, the integration becomes much simpler. The more such questions you solve, the faster you’ll become at spotting the right approach.
If you prefer learning through videos, many of these concepts and questions are also explained on my YouTube Channel @MathsBetter. Watching the solutions alongside practice can help you build speed and confidence.
Now, let’s proceed to the next question.
Question 9: Integrals 7.4
9. Evaluate \displaystyle \int \frac{\sec^2 x}{\sqrt{\tan^2x+4}}\,dx.
Solution
Let, \displaystyle I=\int \frac{\sec^2 x}{\sqrt{\tan^2x+4}}\,dx
Since the numerator contains \sec^2x, which is the derivative of \tan x,
Let t=\tan x
Then \displaystyle dt=\sec^2x\,dx
Substituting, we get
\displaystyle I=\int \frac{dt}{\sqrt{t^2+4}}
\displaystyle I=\int \frac{dt}{\sqrt{t^2+2^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C
Using the above formula with a=2, we get
\displaystyle I=\log\left|t+\sqrt{t^2+4}\right|+C
Substituting t=\tan x,
\displaystyle I=\log\left|\tan x+\sqrt{\tan^2x+4}\right|+C
Hence,
\boxed{\displaystyle I=\log\left|\tan x+\sqrt{\tan^2x+4}\right|+C}
Question 10: Integrals 7.4
10. Evaluate \displaystyle \int \frac{dx}{\sqrt{x^2+2x+2}}.
Solution
Let, \displaystyle I=\int \frac{dx}{\sqrt{x^2+2x+2}}
To complete the square, add and subtract the square of half the coefficient of x.
Since the coefficient of x is 2, we have
\displaystyle \left(\frac{2}{2}\right)^2=1
Adding and subtracting 1, we get
\displaystyle x^2+2x+2=x^2+2x+1-1+2
\displaystyle =x^2+2x+1+1
Using the identity
\displaystyle a^2+2ab+b^2=(a+b)^2
we get
\displaystyle x^2+2x+2=(x+1)^2+1
Hence,
\displaystyle I=\int \frac{dx}{\sqrt{(x+1)^2+1}}
Let t=x+1
Then \displaystyle dt=dx
Substituting, we get
\displaystyle I=\int \frac{dt}{\sqrt{t^2+1}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C
Using the above formula with a=1, we get
\displaystyle I=\log\left|t+\sqrt{t^2+1}\right|+C
Substituting t=x+1,
\displaystyle I=\log\left|x+1+\sqrt{x^2+2x+2}\right|+C
Hence,
\boxed{\displaystyle I=\log\left|x+1+\sqrt{x^2+2x+2}\right|+C}
Question 11: Standard Form of Integral
11. Evaluate \displaystyle \int \frac{dx}{9x^2+6x+5}.
Solution
Let, \displaystyle I=\int \frac{dx}{9x^2+6x+5}
To complete the square, first make the coefficient of x^2 equal to 1.
\displaystyle 9x^2+6x+5=9\left(x^2+\frac23x\right)+5
Now, the coefficient of x is \frac23.
Half of this coefficient is
\displaystyle \frac{\frac23}{2}=\frac13
Adding and subtracting its square i.e. \displaystyle \left(\frac13\right)^2=\frac19 inside the bracket, we get
\displaystyle 9x^2+6x+5=9\left(x^2+\frac23x+\frac19-\frac19\right)+5
\displaystyle =9\left(x^2+\frac23x+\frac19\right)-1+5
Using the identity
\displaystyle a^2+2ab+b^2=(a+b)^2
we get \displaystyle 9x^2+6x+5=9\left(x+\frac13\right)^2+4
\displaystyle =9\left[\left(x+\frac13\right)^2+\frac49\right]
\displaystyle =9\left[\left(x+\frac13\right)^2+\left(\frac23\right)^2\right]
Therefore,
\displaystyle I=\int \frac{dx}{9\left[\left(x+\frac13\right)^2+\left(\frac23\right)^2\right]}
\displaystyle I=\frac19\int \frac{dx}{\left(x+\frac13\right)^2+\left(\frac23\right)^2}
Let \displaystyle t=x+\frac13
Then \displaystyle dt=dx
Substituting, we get
\displaystyle I=\frac19\int \frac{dt}{t^2+\left(\frac23\right)^2}
This is of the type
\displaystyle \int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with \displaystyle a=\frac23, we get
\displaystyle I=\frac19\left(\frac32\right)\tan^{-1}\left(\frac{t}{2/3}\right)+C
\displaystyle I=\frac16\tan^{-1}\left(\frac{3t}{2}\right)+C
Substituting \displaystyle t=x+\frac13,
\displaystyle I=\frac16\tan^{-1}\left(\frac{3\left(x+\frac13\right)}{2}\right)+C
\displaystyle I=\frac16\tan^{-1}\left(\frac{3x+1}{2}\right)+C
Hence,
\boxed{\displaystyle \int \frac{dx}{9x^2+6x+5}=\frac16\tan^{-1}\left(\frac{3x+1}{2}\right)+C}
Question 12: Integrals 7.4
12. Evaluate \displaystyle \int \frac{dx}{\sqrt{7-6x-x^2}}.
Solution
Let, \displaystyle I=\int \frac{dx}{\sqrt{7-6x-x^2}}
To complete the square, first take -1 common from the terms containing x.
\displaystyle 7-6x-x^2=-(x^2+6x)+7
Now, the coefficient of x is 6.
Half of this coefficient is
\displaystyle \frac{6}{2}=3
Adding and subtracting its square i.e. \displaystyle 3^2=9 inside the bracket, we get
\displaystyle 7-6x-x^2=-\left(x^2+6x+9-9\right)+7
\displaystyle =-\left(x^2+6x+9\right)+9+7
Using the identity
\displaystyle a^2+2ab+b^2=(a+b)^2
we get
\displaystyle 7-6x-x^2=-(x+3)^2+16
\displaystyle =4^2-(x+3)^2
Therefore,
\displaystyle I=\int \frac{dx}{\sqrt{4^2-(x+3)^2}}
Let t=x+3
Then dt=dx
Substituting, we get
\displaystyle I=\int \frac{dt}{\sqrt{4^2-t^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with a=4, we get
\displaystyle I=\sin^{-1}\left(\frac{t}{4}\right)+C
Substituting back, t=x+3,
\displaystyle I=\sin^{-1}\left(\frac{x+3}{4}\right)+C
Hence,
\boxed{\displaystyle \int \frac{dx}{\sqrt{7-6x-x^2}}=\sin^{-1}\left(\frac{x+3}{4}\right)+C}
Tip: Before starting the integration, identify which standard form the given integral can be reduced to. A few lines of algebraic simplification at the beginning can often save a lot of work later.
Question 13: Integrals 7.4
13. Evaluate \displaystyle \int \frac{dx}{\sqrt{(x-1)(x-2)}}.
Solution
Let, \displaystyle I=\int \frac{dx}{\sqrt{(x-1)(x-2)}}
First, open the brackets.
\displaystyle (x-1)(x-2)=x^2-3x+2
Therefore,
\displaystyle I=\int \frac{dx}{\sqrt{x^2-3x+2}}
To complete the square, the coefficient of x^2 is already 1.
The coefficient of x is -3.
Half of this coefficient is
\displaystyle \frac{-3}{2}=-\frac32
Adding and subtracting its square i.e. \displaystyle \left(-\frac32\right)^2=\frac94, we get
\displaystyle x^2-3x+2=x^2-3x+\frac94-\frac94+2
\displaystyle =\left(x^2-3x+\frac94\right)-\frac14
Using the identity
\displaystyle a^2-2ab+b^2=(a-b)^2
we get
\displaystyle x^2-3x+2=\left(x-\frac32\right)^2-\left(\frac12\right)^2
Hence,
\displaystyle I=\int \frac{dx}{\sqrt{\left(x-\frac32\right)^2-\left(\frac12\right)^2}}
Let \displaystyle t=x-\frac32
Then \displaystyle dt=dx
Substituting, we get
\displaystyle I=\int \frac{dt}{\sqrt{t^2-\left(\frac12\right)^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C
Using the above formula with \displaystyle a=\frac12, we get
\displaystyle I=\log\left|t+\sqrt{t^2-\frac14}\right|+C
Substituting \displaystyle t=x-\frac32, and simplifying
\displaystyle \Rightarrow I=\log\left|x-\frac32+\sqrt{x^2-3x+2}\right|+C
Hence,
\boxed{\displaystyle I=\log\left|x-\frac32+\sqrt{x^2-3x+2}\right|+C}
Question 14: Integrals Exercise 7.4
14. Evaluate \displaystyle \int \frac{dx}{\sqrt{8+3x-x^2}}.
Solution
Let, \displaystyle I=\int \frac{dx}{\sqrt{8+3x-x^2}}
To complete the square, first take -1 common from the terms containing x.
\displaystyle 8+3x-x^2=-(x^2-3x)+8
Now, the coefficient of x is -3.
Half of this coefficient is
\displaystyle \frac{-3}{2}=-\frac32
Adding and subtracting its square i.e. \displaystyle \left(-\frac32\right)^2=\frac94 inside the bracket,
We get, \displaystyle 8+3x-x^2=-(x^2-3x+\frac94-\frac94)+8
\displaystyle =-(x^2-3x+\frac94)+\frac94+8
\displaystyle =-(x^2-3x+\frac94)+\frac{41}{4}
Using the identity
\displaystyle a^2-2ab+b^2=(a-b)^2
we get
\displaystyle 8+3x-x^2=\frac{41}{4}-\left(x-\frac32\right)^2
\displaystyle =\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac32\right)^2
Therefore,
\displaystyle I=\int \frac{dx}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac32\right)^2}}
Let \displaystyle t=x-\frac32
Then \displaystyle dt=dx
Substituting, we get
\displaystyle I=\int \frac{dt}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-t^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with \displaystyle a=\frac{\sqrt{41}}{2}, we get
\displaystyle I=\sin^{-1}\left(\frac{t}{\sqrt{41}/2}\right)+C
\displaystyle I=\sin^{-1}\left(\frac{2t}{\sqrt{41}}\right)+C
Substituting \displaystyle t=x-\frac32,
\displaystyle I=\sin^{-1}\left(\frac{2\left(x-\frac32\right)}{\sqrt{41}}\right)+C
\displaystyle I=\sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C
Hence,
\boxed{\displaystyle \int \frac{dx}{\sqrt{8+3x-x^2}}=\sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C}
Important: Whenever you need to complete the square, first rewrite the quadratic so that the coefficient of x^2 becomes +1. Then divide the coefficient of x by 2, square it, and add as well as subtract the same quantity. Following these steps carefully makes it much easier to convert the expression into a standard integral form.
Question 15: Indefinite Integrals 7.4
15. Evaluate \displaystyle \int \frac{dx}{\sqrt{(x-a)(x-b)}}.
Solution
Let, \displaystyle I=\int \frac{dx}{\sqrt{(x-a)(x-b)}}
First, open the brackets.
\displaystyle (x-a)(x-b)=x^2-(a+b)x+ab
Therefore,
\displaystyle I=\int \frac{dx}{\sqrt{x^2-(a+b)x+ab}}
To complete the square, the coefficient of x^2 is already 1.
The coefficient of x is -(a+b).
Half of this coefficient is
\displaystyle \frac{-(a+b)}{2}=-\frac{a+b}{2}
Adding and subtracting its square i.e. \displaystyle \left(-\frac{a+b}{2}\right)^2=\frac{(a+b)^2}{4} inside the expression, we get
\displaystyle x^2-(a+b)x+ab=x^2-(a+b)x+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}+ab
\displaystyle =\left(x^2-(a+b)x+\frac{(a+b)^2}{4}\right)-\frac{(a+b)^2}{4}+ab
Using the identity
\displaystyle a^2-2ab+b^2=(a-b)^2
we get \displaystyle x^2-(a+b)x+ab=\left(x-\frac{a+b}{2}\right)^2-\frac{(a+b)^2}{4}+ab
\displaystyle =\left(x-\frac{a+b}{2}\right)^2-\frac{(a+b)^2-4ab}{4}
\displaystyle =\left(x-\frac{a+b}{2}\right)^2-\frac{a^2-2ab+b^2}{4}
On simplification,
\displaystyle =\left(x-\frac{a+b}{2}\right)^2-\frac{(a-b)^2}{4}
\displaystyle =\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2
Therefore,
\displaystyle I=\int \frac{dx}{\sqrt{\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2}}
Let \displaystyle t=x-\frac{a+b}{2}
Then \displaystyle dt=dx
Substituting, we get
\displaystyle I=\int \frac{dt}{\sqrt{t^2-\left(\frac{a-b}{2}\right)^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C
Using the above formula with \displaystyle a=\frac{a-b}{2}, we get
\displaystyle I=\log\left|t+\sqrt{t^2-\left(\frac{a-b}{2}\right)^2}\right|+C
Substituting \displaystyle t=x-\frac{a+b}{2},
\boxed{\displaystyle I=\log\left|x-\frac{a+b}{2}+\sqrt{(x-a)(x-b)}\right|+C}
Question 16: Integrals 7.4
16. Evaluate \displaystyle \int \frac{4x+1}{\sqrt{2x^2+x-3}}\,dx.
Solution
Let, \displaystyle I=\int \frac{4x+1}{\sqrt{2x^2+x-3}}\,dx
Here the numerator 4x+1 is exactly the derivative of the expression inside the square root, since
\displaystyle \frac{d}{dx}(2x^2+x-3)=4x+1
Therefore, let t=2x^2+x-3
Then \displaystyle dt=(4x+1)\,dx
Substituting, we get
\displaystyle I=\int \frac{dt}{\sqrt t}
\displaystyle I=\int t^{-1/2}\,dt
Using the formula
\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad n\ne -1
we get
\displaystyle I=\frac{t^{1/2}}{1/2}+C
\displaystyle I=2\sqrt t+C
Substituting t=2x^2+x-3,
\displaystyle I=2\sqrt{2x^2+x-3}+C
Hence,
\boxed{\displaystyle \int \frac{4x+1}{\sqrt{2x^2+x-3}}\,dx=2\sqrt{2x^2+x-3}+C}
Question 17: Special Integrals 7.4
17. Evaluate \displaystyle \int \frac{x+2}{\sqrt{x^2-1}}\,dx.
Solution
Let, \displaystyle I=\int \frac{x+2}{\sqrt{x^2-1}}\,dx
The numerator is linear and the denominator contains a quadratic expression. Therefore, express the numerator in the form
\displaystyle x+2=A\frac{d}{dx}(x^2-1)+B
Since
\displaystyle \frac{d}{dx}(x^2-1)=2x
we have
\displaystyle x+2=A(2x)+B
or \displaystyle x+2=2Ax+B
Comparing the coefficients of x and the constant terms on both sides, we get
\displaystyle 2A=1,\qquad B=2
\displaystyle \Rightarrow A=\frac12,\qquad B=2
Therefore,
\displaystyle x+2=\frac12(2x)+2
Substituting in the integral,
\displaystyle I=\int \frac{\frac12(2x)+2}{\sqrt{x^2-1}}\,dx
\displaystyle I=\frac12\int \frac{2x}{\sqrt{x^2-1}}\,dx+2\int \frac{dx}{\sqrt{x^2-1}}
Let \displaystyle I=I_1+I_2 …(1)
where
\displaystyle I_1=\frac12\int \frac{2x}{\sqrt{x^2-1}}\,dx
and
\displaystyle I_2=2\int \frac{dx}{\sqrt{x^2-1}}
To evaluate I_1,
Let \displaystyle t=x^2-1
Then \displaystyle dt=2x\,dx
Substituting,
We get, \displaystyle I_1=\frac12\int \frac{dt}{\sqrt t}
\displaystyle I_1=\frac12\int t^{-1/2}\,dt
\displaystyle I_1=\sqrt t
i.e. \displaystyle I_1=\sqrt{x^2-1} …(2)
To evaluate I_2, note that
\displaystyle I_2=2\int \frac{dx}{\sqrt{x^2-1}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C
Using the above formula with a=1, we get
\displaystyle I_2=2\log\left|x+\sqrt{x^2-1}\right| …(3)
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=\sqrt{x^2-1}+2\log\left|x+\sqrt{x^2-1}\right|+C
Hence,
\boxed{\displaystyle I=\sqrt{x^2-1}+2\log\left|x+\sqrt{x^2-1}\right|+C}
Shortcut: To find the constants quickly, try to create the derivative directly instead of comparing coefficients.
Here,
\displaystyle \frac{d}{dx}(x^2-1)=2x
and the numerator is x+2.
Since
\displaystyle \frac12(2x)=x
we write
\displaystyle x+2=\frac12(2x)+2
or
\displaystyle x+2=\frac12\frac{d}{dx}(x^2-1)+2
This immediately gives
\displaystyle A=\frac12,\qquad B=2
This shortcut is especially useful in MCQs, lengthy integrals and Differential Equations.
Question 18: Integration Using Formulas
18. Evaluate \displaystyle \int \frac{5x-2}{1+2x+3x^2}\,dx.
Solution
Let, \displaystyle I=\int \frac{5x-2}{1+2x+3x^2}\,dx
The numerator is linear and the denominator is quadratic. Therefore, express the numerator in the form
\displaystyle 5x-2=A\frac{d}{dx}(1+2x+3x^2)+B
Since
\displaystyle \frac{d}{dx}(1+2x+3x^2)=2+6x
we have
\displaystyle 5x-2=A(6x+2)+B
or \displaystyle 5x-2=6Ax+(2A+B)
Comparing the coefficients of x and the constant terms on both sides,
We get, \displaystyle 6A=5
and \displaystyle 2A+B=-2
\displaystyle \Rightarrow A=\frac56
and \displaystyle B=-2-\frac{10}{6}=-\frac{11}{3}
Therefore,
\displaystyle 5x-2=\frac56(6x+2)-\frac{11}{3}
Substituting in the integral, we get
\displaystyle I=\frac56\int \frac{6x+2}{1+2x+3x^2}\,dx-\frac{11}{3}\int \frac{dx}{1+2x+3x^2}
Let \displaystyle I=I_1-I_2 …(1)
where
\displaystyle I_1=\frac56\int \frac{6x+2}{1+2x+3x^2}\,dx
and
\displaystyle I_2=\frac{11}{3}\int \frac{dx}{1+2x+3x^2}
To evaluate I_1,
Let \displaystyle t=1+2x+3x^2
Then \displaystyle dt=(6x+2)\,dx
Therefore,
\displaystyle I_1=\frac56\int \frac{dt}{t}
\displaystyle I_1=\frac56\log|t|
i.e. \displaystyle I_1=\frac56\log(1+2x+3x^2) …(2)
To evaluate I_2, complete the square in the denominator.
\displaystyle 1+2x+3x^2=3\left(x^2+\frac23x\right)+1
Half of the coefficient of x is
\displaystyle \frac{\frac23}{2}=\frac13
Adding and subtracting its square i.e. \displaystyle \left(\frac13\right)^2=\frac19 inside the bracket,
We get, \displaystyle 1+2x+3x^2=3\left(x^2+\frac23x+\frac19-\frac19\right)+1
\displaystyle =3\left(x+\frac13\right)^2-\frac13+1
\displaystyle =3\left(x+\frac13\right)^2+\frac23
On simplification,
\displaystyle =3\left[\left(x+\frac13\right)^2+\frac29\right]
\displaystyle =3\left[\left(x+\frac13\right)^2+\left(\frac{\sqrt2}{3}\right)^2\right]
Therefore,
\displaystyle I_2=\frac{11}{9}\int \frac{dx}{\left(x+\frac13\right)^2+\left(\frac{\sqrt2}{3}\right)^2}
Let \displaystyle t=x+\frac13
Then \displaystyle dt=dx
Substituting, we get
\displaystyle I_2=\frac{11}{9}\int \frac{dt}{t^2+\left(\frac{\sqrt2}{3}\right)^2}
This is of the type
\displaystyle \int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with \displaystyle a=\frac{\sqrt2}{3}, we get
\displaystyle I_2=\frac{11}{3\sqrt2}\tan^{-1}\left(\frac{3t}{\sqrt2}\right) …(3)
Substituting \displaystyle t=x+\frac13,
\displaystyle I_2=\frac{11}{3\sqrt2}\tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=\frac56\log(1+2x+3x^2)-\frac{11}{3\sqrt2}\tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)+C
Hence,
\boxed{\begin{aligned} I &= \frac56\log(1+2x+3x^2)\\ &\quad-\frac{11}{3\sqrt2}\tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)+C \end{aligned}}
Shortcut: Instead of comparing coefficients, try to create the derivative directly in the numerator and find A and B.
Here,
\displaystyle \frac{d}{dx}(1+2x+3x^2)=2+6x
and the numerator is 5x-2.
Since the coefficient of x in the derivative is 6, multiply it by \frac56 to obtain 5x.
\displaystyle \frac56(6x+2)=5x+\frac53
To make the constant term equal to -2, subtract
\displaystyle 2+\frac53=\frac{11}{3}
Therefore,
\displaystyle 5x-2=\frac56(6x+2)-\frac{11}{3}
or
\displaystyle 5x-2=\frac56\frac{d}{dx}(1+2x+3x^2)-\frac{11}{3}
This immediately gives
\displaystyle A=\frac56,\qquad B=-\frac{11}{3}
This shortcut is very useful in MCQs, lengthy integrals and Differential Equations.
Question 19: Integrals Exercise 7.4
19. Evaluate \displaystyle \int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx.
Solution
Let, \displaystyle I=\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx
First, open the brackets in the denominator.
\displaystyle (x-5)(x-4)=x^2-9x+20
Therefore,
\displaystyle I=\int \frac{6x+7}{\sqrt{x^2-9x+20}}\,dx
Express the numerator in the form
\displaystyle 6x+7=A\frac{d}{dx}(x^2-9x+20)+B
Since
\displaystyle \frac{d}{dx}(x^2-9x+20)=2x-9
we have
\displaystyle 6x+7=A(2x-9)+B
or \displaystyle 6x+7=2Ax+(-9A+B)
Comparing the coefficients of x and the constant terms on both sides, we get
\displaystyle 2A=6
and \displaystyle -9A+B=7
\displaystyle \Rightarrow A=3
and \displaystyle B=34
Therefore,
\displaystyle 6x+7=3(2x-9)+34
Substituting in the integral, we get
\displaystyle I=3\int \frac{2x-9}{\sqrt{x^2-9x+20}}\,dx+34\int \frac{dx}{\sqrt{x^2-9x+20}}
Let \displaystyle I=I_1+I_2 …(1)
where
\displaystyle I_1=3\int \frac{2x-9}{\sqrt{x^2-9x+20}}\,dx
and
\displaystyle I_2=34\int \frac{dx}{\sqrt{x^2-9x+20}}
To evaluate I_1, let
\displaystyle t=x^2-9x+20
Then
\displaystyle dt=(2x-9)\,dx
Substituting,
We get, \displaystyle I_1=3\int \frac{dt}{\sqrt t}
\displaystyle I_1=3\int t^{-1/2}\,dt
\displaystyle I_1=6\sqrt t
i.e. \displaystyle I_1=6\sqrt{x^2-9x+20} …(2)
To evaluate I_2, complete the square in the denominator.
The coefficient of x is -9.
Half of this coefficient is
\displaystyle \frac{-9}{2}=-\frac92
Adding and subtracting its square i.e. \displaystyle \left(-\frac92\right)^2=\frac{81}{4},
We get, \displaystyle x^2-9x+20=x^2-9x+\frac{81}{4}-\frac{81}{4}+20
\displaystyle =\left(x-\frac92\right)^2-\frac14
\displaystyle =\left(x-\frac92\right)^2-\left(\frac12\right)^2
Therefore,
\displaystyle I_2=34\int \frac{dx}{\sqrt{\left(x-\frac92\right)^2-\left(\frac12\right)^2}}
Let \displaystyle u=x-\frac92
Then \displaystyle du=dx
Substituting, we get
\displaystyle I_2=34\int \frac{du}{\sqrt{u^2-\left(\frac12\right)^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C
Using the above formula with \displaystyle a=\frac12, we get
\displaystyle I_2=34\log\left|u+\sqrt{u^2-\frac14}\right| …(3)
Substituting \displaystyle u=x-\frac92,
\displaystyle I_2=34\log\left|x-\frac92+\sqrt{x^2-9x+20}\right|
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=6\sqrt{x^2-9x+20}+34\log\left|x-\frac92+\sqrt{x^2-9x+20}\right|+C
Hence,
\boxed{\begin{aligned} I &= 6\sqrt{x^2-9x+20}\\ &\quad+34\log\left|x-\frac92+\sqrt{x^2-9x+20}\right|+C \end{aligned}}
Question 20: Integrals 7.4
20. Evaluate \displaystyle \int \frac{x+2}{\sqrt{4x-x^2}}\,dx.
Solution
Let, \displaystyle I=\int \frac{x+2}{\sqrt{4x-x^2}}\,dx
Express the numerator in the form
\displaystyle x+2=A\frac{d}{dx}(4x-x^2)+B
Since
\displaystyle \frac{d}{dx}(4x-x^2)=4-2x
we have
\displaystyle x+2=A(4-2x)+B
and \displaystyle x+2=-2Ax+(4A+B)
Comparing the coefficients of x and the constant terms on both sides, we get
\displaystyle -2A=1
and \displaystyle 4A+B=2
We get, \displaystyle A=-\frac12
and \displaystyle B=4
Therefore,
\displaystyle x+2=-\frac12(4-2x)+4
Substituting in the integral, we get
\displaystyle I=-\frac12\int \frac{4-2x}{\sqrt{4x-x^2}}\,dx+4\int \frac{dx}{\sqrt{4x-x^2}}
Let \displaystyle I=I_1+I_2 …(1)
where
\displaystyle I_1=-\frac12\int \frac{4-2x}{\sqrt{4x-x^2}}\,dx
and
\displaystyle I_2=4\int \frac{dx}{\sqrt{4x-x^2}}
To evaluate I_1,
Let \displaystyle t=4x-x^2
Then \displaystyle dt=(4-2x)\,dx
Substituting,
We get, \displaystyle I_1=-\frac12\int \frac{dt}{\sqrt t}
\displaystyle I_1=-\frac12\int t^{-1/2}\,dt
\displaystyle I_1=-\sqrt t
i.e. \displaystyle I_1=-\sqrt{4x-x^2} …(2)
To evaluate I_2, complete the square in the denominator.
\displaystyle 4x-x^2=-(x^2-4x)
The coefficient of x is -4.
Half of this coefficient is
\displaystyle \frac{-4}{2}=-2
Adding and subtracting its square i.e. \displaystyle (-2)^2=4,
We get, \displaystyle 4x-x^2=-(x^2-4x+4-4)
\displaystyle =-(x-2)^2+4
\displaystyle =2^2-(x-2)^2
Therefore,
\displaystyle I_2=4\int \frac{dx}{\sqrt{2^2-(x-2)^2}}
Let \displaystyle u=x-2
Then \displaystyle du=dx
Substituting, we get
\displaystyle I_2=4\int \frac{du}{\sqrt{2^2-u^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with a=2, we get
\displaystyle I_2=4\sin^{-1}\left(\frac{u}{2}\right) …(3)
\displaystyle I_2=4\sin^{-1}\left(\frac{x-2}{2}\right)
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=-\sqrt{4x-x^2}+4\sin^{-1}\left(\frac{x-2}{2}\right)+C
Hence,
\boxed{\displaystyle I=-\sqrt{4x-x^2}+4\sin^{-1}\left(\frac{x-2}{2}\right)+C}
Question 21: Special Types of Integrals 7.4
21. Evaluate \displaystyle \int \frac{x+2}{\sqrt{x^2+2x+3}}\,dx.
Solution
Let, \displaystyle I=\int \frac{x+2}{\sqrt{x^2+2x+3}}\,dx
Express the numerator in the form
\displaystyle x+2=A\frac{d}{dx}(x^2+2x+3)+B
Since
\displaystyle \frac{d}{dx}(x^2+2x+3)=2x+2
we have
\displaystyle x+2=A(2x+2)+B
or \displaystyle x+2=2Ax+(2A+B)
Comparing the coefficients of x and the constant terms on both sides, we get
\displaystyle 2A=1
and \displaystyle 2A+B=2
\displaystyle \Rightarrow A=\frac12,\qquad B=1
Therefore,
\displaystyle x+2=\frac12(2x+2)+1
Substituting in the integral, we get
\displaystyle I=\frac12\int \frac{2x+2}{\sqrt{x^2+2x+3}}\,dx+\int \frac{dx}{\sqrt{x^2+2x+3}}
Let \displaystyle I=I_1+I_2 …(1)
where
\displaystyle I_1=\frac12\int \frac{2x+2}{\sqrt{x^2+2x+3}}\,dx
and
\displaystyle I_2=\int \frac{dx}{\sqrt{x^2+2x+3}}
To evaluate I_1,
Let \displaystyle t=x^2+2x+3
Then \displaystyle dt=(2x+2)\,dx
Substituting, we get
\displaystyle I_1=\frac12\int \frac{dt}{\sqrt t}
or \displaystyle I_1=\frac12\int t^{-1/2}\,dt
\displaystyle I_1=\frac12\cdot\frac{t^{1/2}}{1/2}
\displaystyle I_1=\sqrt t
i.e. \displaystyle I_1=\sqrt{x^2+2x+3} …(2)
To evaluate I_2, complete the square in the denominator.
The coefficient of x is 2.
Half of this coefficient is
\displaystyle \frac{2}{2}=1
Adding and subtracting its square i.e. \displaystyle 1^2=1,
We get, \displaystyle x^2+2x+3=x^2+2x+1-1+3
\displaystyle =(x+1)^2+2
\displaystyle =(x+1)^2+(\sqrt2)^2
Therefore,
\displaystyle I_2=\int \frac{dx}{\sqrt{(x+1)^2+(\sqrt2)^2}}
Let \displaystyle u=x+1
Then \displaystyle du=dx
Substituting, we get
\displaystyle I_2=\int \frac{du}{\sqrt{u^2+(\sqrt2)^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C
Using the above formula with a=\sqrt2, we get
\displaystyle I_2=\log\left|u+\sqrt{u^2+2}\right| …(3)
Substituting u=x+1,
\displaystyle I_2=\log\left|x+1+\sqrt{x^2+2x+3}\right|
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=\sqrt{x^2+2x+3}+\log\left|x+1+\sqrt{x^2+2x+3}\right|+C
Hence,
\boxed{\begin{aligned} I &= \sqrt{x^2+2x+3}\\ &\quad+\log\left|x+1+\sqrt{x^2+2x+3}\right|+C \end{aligned}}
Question 22: Important Integrals 7.4
22. Evaluate \displaystyle \int \frac{x+3}{x^2-2x-5}\,dx.
Solution
Let, \displaystyle I=\int \frac{x+3}{x^2-2x-5}\,dx
Express the numerator in the form
\displaystyle x+3=A\frac{d}{dx}(x^2-2x-5)+B
Since
\displaystyle \frac{d}{dx}(x^2-2x-5)=2x-2
we have
\displaystyle x+3=A(2x-2)+B
or \displaystyle x+3=2Ax+(-2A+B)
Comparing the coefficients of x and the constant terms on both sides, we get
\displaystyle 2A=1
and \displaystyle -2A+B=3
\displaystyle \Rightarrow A=\frac12,\qquad B=4
Therefore,
\displaystyle x+3=\frac12(2x-2)+4
Substituting in the integral, we get
\displaystyle I=\frac12\int \frac{2x-2}{x^2-2x-5}\,dx+4\int \frac{dx}{x^2-2x-5}
Let \displaystyle I=I_1+I_2 …(1)
where
\displaystyle I_1=\frac12\int \frac{2x-2}{x^2-2x-5}\,dx
and
\displaystyle I_2=4\int \frac{dx}{x^2-2x-5}
To evaluate I_1,
Let \displaystyle t=x^2-2x-5
Then \displaystyle dt=(2x-2)\,dx
Substituting, we get
\displaystyle I_1=\frac12\int \frac{dt}{t}
\displaystyle I_1=\frac12\log|t|
i.e. \displaystyle I_1=\frac12\log|x^2-2x-5| …(2)
To evaluate I_2, complete the square in the denominator.
The coefficient of x is -2.
Half of this coefficient is
\displaystyle \frac{-2}{2}=-1
Adding and subtracting its square i.e. \displaystyle (-1)^2=1,
We get, \displaystyle x^2-2x-5=x^2-2x+1-1-5
\displaystyle =(x-1)^2-6
\displaystyle =(x-1)^2-(\sqrt6)^2
Therefore,
\displaystyle I_2=4\int \frac{dx}{(x-1)^2-(\sqrt6)^2}
Let \displaystyle u=x-1
Then \displaystyle du=dx
Substituting, we get
\displaystyle I_2=4\int \frac{du}{u^2-(\sqrt6)^2}
This is of the type
\displaystyle \int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C
Using the above formula with a=\sqrt6, we get
\displaystyle I_2=\frac{2}{\sqrt6}\log\left|\frac{u-\sqrt6}{u+\sqrt6}\right| …(3)
Substituting u=x-1,
\displaystyle I_2=\frac{2}{\sqrt6}\log\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right|
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=\frac12\log|x^2-2x-5|+\frac{2}{\sqrt6}\log\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right|+C
Hence,
\boxed{\begin{aligned} I &=\frac12\log|x^2-2x-5|\\ &\quad+\frac{2}{\sqrt6}\log\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right|+C \end{aligned}}
Question 23: Integration Using Formulas
23. Evaluate \displaystyle \int \frac{5x+3}{\sqrt{x^2+4x+10}}\,dx.
Solution
Let, \displaystyle I=\int \frac{5x+3}{\sqrt{x^2+4x+10}}\,dx
Express the numerator in the form
\displaystyle 5x+3=A\frac{d}{dx}(x^2+4x+10)+B
Since
\displaystyle \frac{d}{dx}(x^2+4x+10)=2x+4
we have
\displaystyle 5x+3=A(2x+4)+B
or \displaystyle 5x+3=2Ax+(4A+B)
Comparing the coefficients of x and the constant terms on both sides, we get
\displaystyle 2A=5
and \displaystyle 4A+B=3
We get, \displaystyle A=\frac52,\qquad B=-7
Therefore,
\displaystyle 5x+3=\frac52(2x+4)-7
Substituting in the integral, we get
\displaystyle I=\frac52\int \frac{2x+4}{\sqrt{x^2+4x+10}}\,dx-7\int \frac{dx}{\sqrt{x^2+4x+10}}
Let \displaystyle I=I_1-I_2 …(1)
where
\displaystyle I_1=\frac52\int \frac{2x+4}{\sqrt{x^2+4x+10}}\,dx
and
\displaystyle I_2=7\int \frac{dx}{\sqrt{x^2+4x+10}}
To evaluate I_1,
Let \displaystyle t=x^2+4x+10
Then \displaystyle dt=(2x+4)\,dx
Substituting,
We get, \displaystyle I_1=\frac52\int \frac{dt}{\sqrt t}
\displaystyle I_1=\frac52\int t^{-1/2}\,dt
\displaystyle I_1=5\sqrt t
i.e. \displaystyle I_1=5\sqrt{x^2+4x+10} …(2)
To evaluate I_2, complete the square in the denominator.
The coefficient of x is 4.
Half of this coefficient is
\displaystyle \frac{4}{2}=2
Adding and subtracting its square i.e. \displaystyle 2^2=4,
We get, \displaystyle x^2+4x+10=x^2+4x+4-4+10
\displaystyle =(x+2)^2+6
\displaystyle =(x+2)^2+(\sqrt6)^2
Therefore,
\displaystyle I_2=7\int \frac{dx}{\sqrt{(x+2)^2+(\sqrt6)^2}}
Let \displaystyle u=x+2
Then \displaystyle du=dx
Substituting, we get
\displaystyle I_2=7\int \frac{du}{\sqrt{u^2+(\sqrt6)^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C
Using the above formula with a=\sqrt6, we get
\displaystyle I_2=7\log\left|u+\sqrt{u^2+6}\right| …(3)
Substituting u=x+2,
\displaystyle I_2=7\log\left|x+2+\sqrt{x^2+4x+10}\right|
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=5\sqrt{x^2+4x+10}-7\log\left|x+2+\sqrt{x^2+4x+10}\right|+C
Hence,
\boxed{\begin{aligned} I &= 5\sqrt{x^2+4x+10}\\ &\quad-7\log\left|x+2+\sqrt{x^2+4x+10}\right|+C \end{aligned}}
Question 24: Integrals 7.4 – MCQ
24. Choose the correct answer.
\displaystyle \int \frac{dx}{x^2+2x+2}
- (A) x\tan^{-1}(x+1)+C
- (B) \tan^{-1}(x+1)+C
- (C) (x+1)\tan^{-1}x+C
- (D) \tan^{-1}x+C
Solution
Complete the square in the denominator:
\displaystyle x^2+2x+2=x^2+2x+1-1+2
\displaystyle =(x+1)^2+1
Therefore,
\displaystyle \int \frac{dx}{x^2+2x+2}=\int \frac{dx}{(x+1)^2+1}
Let t=x+1
Then dt=dx
Substituting, we get
\displaystyle \int \frac{dt}{t^2+1}
This is of the type
\displaystyle \int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with a=1, we get
\displaystyle \int \frac{dx}{x^2+2x+2}=\tan^{-1}(x+1)+C
✅️ Hence, the correct answer is (B).
Question 25: Integrals 7.4 – MCQ
25. Choose the correct answer.
\displaystyle \int \frac{dx}{\sqrt{9x-4x^2}}
- (A) \displaystyle \frac19\sin^{-1}\left(\frac{9x-8}{8}\right)+C
- (B) \displaystyle \frac12\sin^{-1}\left(\frac{8x-9}{9}\right)+C
- (C) \displaystyle \frac13\sin^{-1}\left(\frac{9x-8}{8}\right)+C
- (D) \displaystyle \frac12\sin^{-1}\left(\frac{9x-8}{9}\right)+C
Solution
Complete the square inside the square root:
\displaystyle 9x-4x^2=-4\left(x^2-\frac94x\right)
Half of the coefficient of x inside the bracket is
\displaystyle \frac{-9/4}{2}=-\frac98
Adding and subtracting its square i.e. \displaystyle \left(\frac98\right)^2=\frac{81}{64} inside the bracket,
We get, \displaystyle 9x-4x^2=-4\left(x^2-\frac94x+\frac{81}{64}-\frac{81}{64}\right)
\displaystyle =-4\left(x-\frac98\right)^2+\frac{81}{16}
\displaystyle =\left(\frac94\right)^2-\left(2x-\frac94\right)^2
Therefore,
\displaystyle I=\int \frac{dx}{\sqrt{\left(\frac94\right)^2-\left(2x-\frac94\right)^2}}
Let \displaystyle t=2x-\frac94
Then \displaystyle dt=2\,dx
or \displaystyle dx=\frac12\,dt
Substituting, we get
\displaystyle I=\frac12\int \frac{dt}{\sqrt{\left(\frac94\right)^2-t^2}}
This is of the type
\displaystyle \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with \displaystyle a=\frac94, we get
\displaystyle I=\frac12\sin^{-1}\left(\frac{t}{9/4}\right)+C
\displaystyle =\frac12\sin^{-1}\left(\frac{8x-9}{9}\right)+C
✅️ Hence, the correct answer is (B).
Common Mistakes to Avoid
- Not identifying the standard form correctly: Before integrating, rewrite the given integral so that it matches one of the standard forms involving a^2-x^2, a^2+x^2 or x^2-a^2.
- Skipping the completion of the square: For expressions such as ax^2+bx+c, complete the square carefully. Most mistakes occur while converting the quadratic into a standard form.
- Making errors while completing the square: Always divide the coefficient of x by 2, square it, and then add as well as subtract the same quantity. Do not jump directly to the final result.
- Choosing the wrong substitution: Use x=a\sin\theta for a^2-x^2, x=a\tan\theta for a^2+x^2, and x=a\sec\theta for x^2-a^2.
- Ignoring the derivative of the denominator: In integrals of the form \displaystyle \int \frac{px+q}{ax^2+bx+c}\,dx or \displaystyle \int \frac{px+q}{\sqrt{ax^2+bx+c}}\,dx, first express the numerator as A\left(\frac{d}{dx}\text{(denominator)}\right)+B.
- Not splitting the integral after finding A and B: After expressing the numerator appropriately, split the integral into I_1 and I_2 and integrate each part separately.
- Using the wrong standard formula: Similar-looking formulas involving \sin^{-1}x, \tan^{-1}x and logarithms are often confused. Verify the denominator carefully before applying a formula.
- Forgetting the substitution back in terms of x: After integrating in terms of t or u, always substitute back to obtain the final answer in terms of the original variable.
- Making algebraic mistakes while simplifying logarithmic answers: Be careful while simplifying expressions inside logarithms and while handling fractions or surds.
- Forgetting the constant of integration: Every indefinite integral must end with C.
Continue Learning
After completing Exercise 7.4 of Integrals, you should now be comfortable with evaluating a variety of special integrals by reducing them to standard forms. In this exercise, you learned how to use suitable substitutions, complete the square in quadratic expressions and apply standard integration formulas involving inverse trigonometric and logarithmic functions.
To strengthen your understanding further, make sure that you revise:
- The standard integral forms involving a^2-x^2, a^2+x^2 and x^2-a^2
- The corresponding substitutions x=a\sin\theta, x=a\tan\theta and x=a\sec\theta
- Completing the square in quadratic expressions of the form ax^2+bx+c
- Reducing quadratic expressions to one of the standard integral forms before integration
- Integrals involving \displaystyle \frac{1}{ax^2+bx+c} and \displaystyle \frac{1}{\sqrt{ax^2+bx+c}}
- Expressing the numerator as \displaystyle A\left(\frac{d}{dx}\text{(denominator)}\right)+B for linear-over-quadratic and linear-over-root-quadratic integrals
- Splitting integrals into simpler parts such as I_1 and I_2 and evaluating each separately
- Standard formulas involving \sin^{-1}x, \tan^{-1}x and logarithmic functions
- Substituting back in terms of the original variable after integration
- Verifying that the final answer is written in its simplest form along with the constant of integration C
Explore More
Special integrals become much easier once you learn to recognize the underlying pattern. As you practise more questions, focus on reducing the given integral to a standard form, completing the square carefully, choosing the correct substitution and applying the appropriate standard formula. Also remember to express the numerator in terms of the derivative of the denominator whenever possible, simplify the final answer and include the constant of integration C. With regular practice, these techniques will become natural and help you solve even complicated integrals with confidence.
All the best and keep learning 👍
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