Complete solutions for Class 12 Maths (NCERT) Exercise 7.4 on Integrals. Learn to apply the six standard formulas for integration involving $\mathbf{x^2 \pm a^2}$ and $\mathbf{\sqrt{x^2 \pm a^2}}$. Covers the crucial technique of completing the square and integrating types like $\int \frac{px+q}{\text{quadratic}} dx$.
Table of Contents
This exercise focuses on using the six standard integration formulas involving $\mathbf{x^2 \pm a^2}$ in the denominator, often requiring completing the square or substitution.
Part 1: Direct Application of Standard Formulas (1-15)
The six key formulas used in this section are:
- $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$
- $\int \frac{dx}{x^2 – a^2} = \frac{1}{2a} \log\left|\frac{x – a}{x + a}\right| + C$
- $\int \frac{dx}{a^2 – x^2} = \frac{1}{2a} \log\left|\frac{a + x}{a – x}\right| + C$
- $\int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C$
- $\int \frac{dx}{\sqrt{x^2 – a^2}} = \log\left|x + \sqrt{x^2 – a^2}\right| + C$
- $\int \frac{dx}{\sqrt{a^2 – x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$
1. $\int \frac{3x^2}{x^6 + 1} dx$
Let $u = x^3$. Then $du = 3x^2 dx$.
$$\int \frac{du}{u^2 + 1} = \tan^{-1}(u) + C = \mathbf{\tan^{-1}(x^3) + C}$$
2. $\int \frac{1}{\sqrt{1 + 4x^2}} dx$
Factor out 4 from under the root: $\int \frac{dx}{\sqrt{4(1/4 + x^2)}} = \int \frac{dx}{2\sqrt{(1/2)^2 + x^2}}$.
Using formula 4 with $a = 1/2$:
$$\frac{1}{2} \int \frac{dx}{\sqrt{x^2 + (1/2)^2}} = \frac{1}{2} \log\left|x + \sqrt{x^2 + \frac{1}{4}}\right| + C$$
$$\frac{1}{2} \log\left|x + \frac{\sqrt{4x^2 + 1}}{2}\right| + C = \mathbf{\frac{1}{2} \log|2x + \sqrt{4x^2 + 1}| + C}$$
3. $\int \frac{1}{\sqrt{(2 – x)^2 + 1}} dx$
Let $u = 2 – x$. Then $du = -dx$.
$$\int \frac{-du}{\sqrt{u^2 + 1^2}} = -\log|u + \sqrt{u^2 + 1}| + C$$
$$= \mathbf{-\log|2 – x + \sqrt{(2 – x)^2 + 1}| + C}$$
4. $\int \frac{1}{\sqrt{9 – 25x^2}} dx$
Factor out 25 from under the root: $\int \frac{dx}{\sqrt{25(9/25 – x^2)}} = \int \frac{dx}{5\sqrt{(3/5)^2 – x^2}}$.
Using formula 6 with $a = 3/5$:
$$\frac{1}{5} \int \frac{dx}{\sqrt{(3/5)^2 – x^2}} = \frac{1}{5} \sin^{-1}\left(\frac{x}{3/5}\right) + C$$
$$= \mathbf{\frac{1}{5} \sin^{-1}\left(\frac{5x}{3}\right) + C}$$
5. $\int \frac{3x}{1 + 2x^4} dx$
Let $u = \sqrt{2} x^2$. Then $u^2 = 2x^4$.
$du = 2\sqrt{2} x dx$, so $x dx = \frac{1}{2\sqrt{2}} du$.
$$\int \frac{3}{1 + 2x^4} x dx = 3 \int \frac{1}{1 + u^2} \frac{1}{2\sqrt{2}} du$$
$$= \frac{3}{2\sqrt{2}} \tan^{-1}(u) + C = \mathbf{\frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2} x^2) + C}$$
6. $\int \frac{x^2}{1 – x^6} dx$
Let $u = x^3$. Then $du = 3x^2 dx$, so $x^2 dx = \frac{1}{3} du$.
$$\int \frac{1}{1 – u^2} \frac{1}{3} du = \frac{1}{3} \int \frac{du}{1^2 – u^2}$$
Using formula 3 with $a = 1$:
$$= \frac{1}{3} \left[ \frac{1}{2(1)} \log\left|\frac{1 + u}{1 – u}\right| \right] + C = \mathbf{\frac{1}{6} \log\left|\frac{1 + x^3}{1 – x^3}\right| + C}$$
7. $\int \frac{x – 1}{\sqrt{x^2 – 1}} dx$
Separate the fraction:
$$\int \frac{x}{\sqrt{x^2 – 1}} dx – \int \frac{1}{\sqrt{x^2 – 1}} dx$$
- Term 1: Let $u = x^2 – 1$, $du = 2x dx$. $\frac{1}{2} \int u^{-1/2} du = u^{1/2} = \sqrt{x^2 – 1}$.
- Term 2: Standard form (Formula 5 with $a=1$).$$= \mathbf{\sqrt{x^2 – 1} – \log|x + \sqrt{x^2 – 1}| + C}$$
8. $\int \frac{x^2}{\sqrt{x^6 + a^6}} dx$
Let $u = x^3$. Then $du = 3x^2 dx$, so $x^2 dx = \frac{1}{3} du$. The denominator is $\sqrt{(x^3)^2 + (a^3)^2} = \sqrt{u^2 + (a^3)^2}$.
Using formula 4 with $A = a^3$:
$$\frac{1}{3} \int \frac{du}{\sqrt{u^2 + (a^3)^2}} = \frac{1}{3} \log|u + \sqrt{u^2 + a^6}| + C$$
$$= \mathbf{\frac{1}{3} \log|x^3 + \sqrt{x^6 + a^6}| + C}$$
9. $\int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} dx$
Let $u = \tan x$. Then $du = \sec^2 x dx$.
$$\int \frac{du}{\sqrt{u^2 + 2^2}}$$
Using formula 4 with $a = 2$:
$$= \log|u + \sqrt{u^2 + 4}| + C = \mathbf{\log|\tan x + \sqrt{\tan^2 x + 4}| + C}$$
10. $\int \frac{1}{\sqrt{x^2 + 2x + 2}} dx$
Complete the square in the denominator: $x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1^2$.
Let $u = x + 1$. Then $du = dx$.
$$\int \frac{du}{\sqrt{u^2 + 1^2}}$$
Using formula 4 with $a = 1$:
$$= \log|u + \sqrt{u^2 + 1}| + C = \mathbf{\log|x + 1 + \sqrt{x^2 + 2x + 2}| + C}$$
11. $\int \frac{1}{9x^2 + 6x + 5} dx$
Factor out 9: $\int \frac{1}{9(x^2 + \frac{6}{9}x + \frac{5}{9})} dx = \frac{1}{9} \int \frac{1}{x^2 + \frac{2}{3}x + \frac{5}{9}} dx$.
Complete the square: $x^2 + \frac{2}{3}x + \left(\frac{1}{3}\right)^2 – \left(\frac{1}{3}\right)^2 + \frac{5}{9} = \left(x + \frac{1}{3}\right)^2 + \frac{4}{9} = \left(x + \frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2$.
Let $u = x + 1/3$. Then $du = dx$.
$$\frac{1}{9} \int \frac{du}{u^2 + (2/3)^2}$$
Using formula 1 with $a = 2/3$:
$$= \frac{1}{9} \left[ \frac{1}{2/3} \tan^{-1}\left(\frac{u}{2/3}\right) \right] + C = \frac{1}{9} \left[\frac{3}{2} \tan^{-1}\left(\frac{3u}{2}\right)\right] + C$$
$$= \mathbf{\frac{1}{6} \tan^{-1}\left(\frac{3x + 1}{2}\right) + C}$$
12. $\int \frac{1}{\sqrt{7 – 6x – x^2}} dx$
Complete the square: $7 – (x^2 + 6x) = 7 – (x^2 + 6x + 9 – 9) = 7 – (x + 3)^2 + 9 = 16 – (x + 3)^2 = 4^2 – (x + 3)^2$.
Let $u = x + 3$. Then $du = dx$.
$$\int \frac{du}{\sqrt{4^2 – u^2}}$$
Using formula 6 with $a = 4$:
$$= \sin^{-1}\left(\frac{u}{4}\right) + C = \mathbf{\sin^{-1}\left(\frac{x + 3}{4}\right) + C}$$
13. $\int \frac{1}{\sqrt{(x – 1)(x – 2)}} dx$
Multiply the terms: $(x – 1)(x – 2) = x^2 – 3x + 2$.
Complete the square: $x^2 – 3x + \left(\frac{3}{2}\right)^2 – \left(\frac{3}{2}\right)^2 + 2 = \left(x – \frac{3}{2}\right)^2 – \frac{9}{4} + \frac{8}{4} = \left(x – \frac{3}{2}\right)^2 – \left(\frac{1}{2}\right)^2$.
Let $u = x – 3/2$. Then $du = dx$.
$$\int \frac{du}{\sqrt{u^2 – (1/2)^2}}$$
Using formula 5 with $a = 1/2$:
$$= \log\left|u + \sqrt{u^2 – \frac{1}{4}}\right| + C = \mathbf{\log\left|x – \frac{3}{2} + \sqrt{x^2 – 3x + 2}\right| + C}$$
14. $\int \frac{1}{\sqrt{8 + 3x – x^2}} dx$
Complete the square: $8 – (x^2 – 3x) = 8 – (x^2 – 3x + 9/4 – 9/4) = 8 + 9/4 – (x – 3/2)^2 = 41/4 – (x – 3/2)^2 = \left(\frac{\sqrt{41}}{2}\right)^2 – \left(x – \frac{3}{2}\right)^2$.
Let $u = x – 3/2$. Then $du = dx$.
$$\int \frac{du}{\sqrt{(\sqrt{41}/2)^2 – u^2}}$$
Using formula 6 with $a = \sqrt{41}/2$:
$$= \sin^{-1}\left(\frac{u}{\sqrt{41}/2}\right) + C = \mathbf{\sin^{-1}\left(\frac{2x – 3}{\sqrt{41}}\right) + C}$$
15. $\int \frac{1}{\sqrt{(x – a)(x – b)}} dx$
Multiply the terms: $(x – a)(x – b) = x^2 – (a + b)x + ab$.
Complete the square: $\left(x – \frac{a + b}{2}\right)^2 – \left(\frac{a + b}{2}\right)^2 + ab = \left(x – \frac{a + b}{2}\right)^2 – \frac{a^2 + 2ab + b^2 – 4ab}{4} = \left(x – \frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2$.
Let $u = x – \frac{a + b}{2}$. Then $du = dx$.
Using formula 5 with $A = \frac{a – b}{2}$:
$$= \log\left|u + \sqrt{u^2 – A^2}\right| + C = \mathbf{\log\left|x – \frac{a + b}{2} + \sqrt{(x – a)(x – b)}\right| + C}$$
Part 2: Integration of the form $\int \frac{px + q}{\text{quadratic}} dx$ (16-23)
Use the method: $\mathbf{px + q = A \cdot \frac{d}{dx}(\text{quadratic}) + B}$.
16. $\int \frac{4x + 1}{2x^2 + x – 3} dx$
Let $u = 2x^2 + x – 3$. Then $du = (4x + 1) dx$.
This is a direct substitution case ($\int \frac{du}{u}$):
$$\int \frac{du}{u} = \log|u| + C = \mathbf{\log|2x^2 + x – 3| + C}$$
17. $\int \frac{x + 2}{\sqrt{x^2 – 1}} dx$
Separate the fraction: $\int \frac{x}{\sqrt{x^2 – 1}} dx + \int \frac{2}{\sqrt{x^2 – 1}} dx$.
- Term 1: Substitution ($u = x^2 – 1$), result is $\sqrt{x^2 – 1}$.
- Term 2: Standard form (Formula 5).$$= \mathbf{\sqrt{x^2 – 1} + 2 \log|x + \sqrt{x^2 – 1}| + C}$$
18. $\int \frac{5x – 2}{1 + 2x + 3x^2} dx$
Let $D = 1 + 2x + 3x^2$. $\frac{dD}{dx} = 2 + 6x$.
$$5x – 2 = A(6x + 2) + B$$
Comparing coefficients:
- $x$ terms: $6A = 5 \implies A = 5/6$.
- Constant terms: $2A + B = -2 \implies 2(5/6) + B = -2 \implies 5/3 + B = -2 \implies B = -2 – 5/3 = -11/3$.The integral is: $\int \frac{\frac{5}{6}(6x + 2) – \frac{11}{3}}{3x^2 + 2x + 1} dx = \frac{5}{6} \int \frac{6x + 2}{3x^2 + 2x + 1} dx – \frac{11}{3} \int \frac{1}{3x^2 + 2x + 1} dx$.
- Term 1: $\frac{5}{6} \log|3x^2 + 2x + 1|$.
- Term 2: Complete the square for $3x^2 + 2x + 1 = 3(x^2 + \frac{2}{3}x + \frac{1}{3}) = 3[(x + 1/3)^2 + 2/9] = 3[(x + 1/3)^2 + (\sqrt{2}/3)^2]$.$$\int \frac{1}{3[(\dots)]} dx = \frac{1}{3} \cdot \frac{1}{\sqrt{2}/3} \tan^{-1}\left(\frac{x + 1/3}{\sqrt{2}/3}\right) = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{3x + 1}{\sqrt{2}}\right)$$
- Final Result: $\mathbf{\frac{5}{6} \log|3x^2 + 2x + 1| – \frac{11}{3\sqrt{2}} \tan^{-1}\left(\frac{3x + 1}{\sqrt{2}}\right) + C}$
19. $\int \frac{6x + 7}{\sqrt{(x – 5)(x – 4)}} dx$
Denominator: $\sqrt{x^2 – 9x + 20}$. $\frac{d}{dx} (\dots) = 2x – 9$.
$$6x + 7 = A(2x – 9) + B$$
- $x$ terms: $2A = 6 \implies A = 3$.
- Constant terms: $-9A + B = 7 \implies -9(3) + B = 7 \implies B = 34$.The integral is: $\int \frac{3(2x – 9) + 34}{\sqrt{x^2 – 9x + 20}} dx = 3 \int \frac{2x – 9}{\sqrt{x^2 – 9x + 20}} dx + 34 \int \frac{1}{\sqrt{x^2 – 9x + 20}} dx$.
- Term 1: $3 \cdot (2 \sqrt{x^2 – 9x + 20}) = 6 \sqrt{x^2 – 9x + 20}$.
- Term 2: Complete the square: $x^2 – 9x + 20 = (x – 9/2)^2 – 81/4 + 80/4 = (x – 9/2)^2 – (1/2)^2$.$$34 \log\left|x – \frac{9}{2} + \sqrt{x^2 – 9x + 20}\right|$$
- Final Result: $\mathbf{6 \sqrt{x^2 – 9x + 20} + 34 \log\left|x – \frac{9}{2} + \sqrt{x^2 – 9x + 20}\right| + C}$
20. $\int \frac{x + 2}{\sqrt{4x – x^2}} dx$
Denominator: $\sqrt{4x – x^2}$. $\frac{d}{dx} (\dots) = 4 – 2x$.
$$x + 2 = A(4 – 2x) + B$$
- $x$ terms: $-2A = 1 \implies A = -1/2$.
- Constant terms: $4A + B = 2 \implies 4(-1/2) + B = 2 \implies -2 + B = 2 \implies B = 4$.The integral is: $\int \frac{-1/2(4 – 2x) + 4}{\sqrt{4x – x^2}} dx = -\frac{1}{2} \int \frac{4 – 2x}{\sqrt{4x – x^2}} dx + 4 \int \frac{1}{\sqrt{4x – x^2}} dx$.
- Term 1: $-\frac{1}{2} \cdot (2 \sqrt{4x – x^2}) = -\sqrt{4x – x^2}$.
- Term 2: Complete the square: $-(x^2 – 4x) = -(x^2 – 4x + 4 – 4) = 4 – (x – 2)^2 = 2^2 – (x – 2)^2$.$$4 \sin^{-1}\left(\frac{x – 2}{2}\right)$$
- Final Result: $\mathbf{-\sqrt{4x – x^2} + 4 \sin^{-1}\left(\frac{x – 2}{2}\right) + C}$
21. $\int \frac{x + 2}{\sqrt{x^2 + 2x + 3}} dx$
Denominator: $\sqrt{x^2 + 2x + 3}$. $\frac{d}{dx} (\dots) = 2x + 2$.
$$x + 2 = A(2x + 2) + B$$
- $x$ terms: $2A = 1 \implies A = 1/2$.
- Constant terms: $2A + B = 2 \implies 2(1/2) + B = 2 \implies 1 + B = 2 \implies B = 1$.The integral is: $\int \frac{1/2(2x + 2) + 1}{\sqrt{x^2 + 2x + 3}} dx = \frac{1}{2} \int \frac{2x + 2}{\sqrt{x^2 + 2x + 3}} dx + \int \frac{1}{\sqrt{x^2 + 2x + 3}} dx$.
- Term 1: $\frac{1}{2} \cdot (2 \sqrt{x^2 + 2x + 3}) = \sqrt{x^2 + 2x + 3}$.
- Term 2: Complete the square: $x^2 + 2x + 3 = (x + 1)^2 + 2 = (x + 1)^2 + (\sqrt{2})^2$.$$\log|x + 1 + \sqrt{x^2 + 2x + 3}|$$
- Final Result: $\mathbf{\sqrt{x^2 + 2x + 3} + \log|x + 1 + \sqrt{x^2 + 2x + 3}| + C}$
22. $\int \frac{x + 3}{x^2 – 2x – 5} dx$
Denominator: $D = x^2 – 2x – 5$. $\frac{dD}{dx} = 2x – 2$.
$$x + 3 = A(2x – 2) + B$$
- $x$ terms: $2A = 1 \implies A = 1/2$.
- Constant terms: $-2A + B = 3 \implies -2(1/2) + B = 3 \implies -1 + B = 3 \implies B = 4$.The integral is: $\int \frac{1/2(2x – 2) + 4}{x^2 – 2x – 5} dx = \frac{1}{2} \int \frac{2x – 2}{x^2 – 2x – 5} dx + 4 \int \frac{1}{x^2 – 2x – 5} dx$.
- Term 1: $\frac{1}{2} \log|x^2 – 2x – 5|$.
- Term 2: Complete the square: $x^2 – 2x – 5 = (x – 1)^2 – 1 – 5 = (x – 1)^2 – 6 = (x – 1)^2 – (\sqrt{6})^2$.$$4 \cdot \frac{1}{2\sqrt{6}} \log\left|\frac{x – 1 – \sqrt{6}}{x – 1 + \sqrt{6}}\right|$$
- Final Result: $\mathbf{\frac{1}{2} \log|x^2 – 2x – 5| + \frac{2}{\sqrt{6}} \log\left|\frac{x – 1 – \sqrt{6}}{x – 1 + \sqrt{6}}\right| + C}$
23. $\int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} dx$
Denominator: $\sqrt{x^2 + 4x + 10}$. $\frac{d}{dx} (\dots) = 2x + 4$.
$$5x + 3 = A(2x + 4) + B$$
- $x$ terms: $2A = 5 \implies A = 5/2$.
- Constant terms: $4A + B = 3 \implies 4(5/2) + B = 3 \implies 10 + B = 3 \implies B = -7$.The integral is: $\frac{5}{2} \int \frac{2x + 4}{\sqrt{x^2 + 4x + 10}} dx – 7 \int \frac{1}{\sqrt{x^2 + 4x + 10}} dx$.
- Term 1: $\frac{5}{2} \cdot (2 \sqrt{x^2 + 4x + 10}) = 5 \sqrt{x^2 + 4x + 10}$.
- Term 2: Complete the square: $x^2 + 4x + 10 = (x + 2)^2 + 6 = (x + 2)^2 + (\sqrt{6})^2$.$$7 \log|x + 2 + \sqrt{x^2 + 4x + 10}|$$
- Final Result: $\mathbf{5 \sqrt{x^2 + 4x + 10} – 7 \log|x + 2 + \sqrt{x^2 + 4x + 10}| + C}$
Part 3: Multiple Choice Questions
24. $\int \frac{dx}{x^2 + 2x + 2}$
Complete the square: $x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1^2$.
Let $u = x + 1$.
$$\int \frac{du}{u^2 + 1^2} = \frac{1}{1} \tan^{-1}\left(\frac{u}{1}\right) + C = \tan^{-1}(x + 1) + C$$
The correct answer is (B) $\tan^{-1} (x + 1) + C$.
25. $\int \frac{dx}{\sqrt{9x – 4x^2}}$
Factor out the $-4$: $\int \frac{dx}{\sqrt{-4(x^2 – 9/4 x)}} = \int \frac{dx}{2\sqrt{-(x^2 – 9/4 x)}}$.
Complete the square: $x^2 – \frac{9}{4}x = \left(x – \frac{9}{8}\right)^2 – \left(\frac{9}{8}\right)^2$.
$$2\sqrt{-\left[\left(x – \frac{9}{8}\right)^2 – \left(\frac{9}{8}\right)^2\right]} = 2\sqrt{\left(\frac{9}{8}\right)^2 – \left(x – \frac{9}{8}\right)^2}$$
Let $u = x – 9/8$. $a = 9/8$.
$$\frac{1}{2} \int \frac{du}{\sqrt{a^2 – u^2}} = \frac{1}{2} \sin^{-1}\left(\frac{u}{a}\right) + C$$
$$= \frac{1}{2} \sin^{-1}\left(\frac{x – 9/8}{9/8}\right) + C = \mathbf{\frac{1}{2} \sin^{-1}\left(\frac{8x – 9}{9}\right) + C}$$
The correct answer is (D) $\frac{1}{2} \sin^{-1}\left(\frac{8x – 9}{9}\right) + C$.