Integrals Class 12 Exercise 7.9 Solutions | Mathematics Part II CBSE

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• Ex. 7.1


• Ex. 7.2


• Ex. 7.2


• Ex. 7.3


• Ex. 7.4


• Ex. 7.5


• Ex. 7.6


• Ex. 7.7


• Ex. 7.8


• Ex. 7.9


• Ex. 7.10


• Ex. 7.11


• Miscellaneous Exercise

Exercise 7.9 Solutions

Topic: Definite Integrals
Evaluate the following definite integrals:

Question 1. Evaluate \(\int_{-1}^{1} (x+1) dx\)

Let \(I = \int_{-1}^{1} (x+1) dx\)

Integration of \(x^n\) is \(\frac{x^{n+1}}{n+1}\).

$$I = \left[ \frac{x^2}{2} + x \right]_{-1}^{1}$$

Applying limits (Upper Limit - Lower Limit):

$$I = \left( \frac{1^2}{2} + 1 \right) - \left( \frac{(-1)^2}{2} + (-1) \right)$$ $$I = \left( \frac{1}{2} + 1 \right) - \left( \frac{1}{2} - 1 \right)$$ $$I = \frac{3}{2} - \left( -\frac{1}{2} \right)$$ $$I = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$$

Answer: 2

Question 2. Evaluate \(\int_{2}^{3} \frac{1}{x} dx\)

Let \(I = \int_{2}^{3} \frac{1}{x} dx\)

We know that \(\int \frac{1}{x} dx = \log |x|\).

$$I = \left[ \log |x| \right]_{2}^{3}$$

Applying limits:

$$I = \log 3 - \log 2$$ $$I = \log \left( \frac{3}{2} \right)$$

Answer: \(\log \frac{3}{2}\)

Question 3. Evaluate \(\int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) dx\)

Let \(I = \int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) dx\)

Integrating term by term:

$$I = \left[ 4\frac{x^4}{4} - 5\frac{x^3}{3} + 6\frac{x^2}{2} + 9x \right]_{1}^{2}$$ $$I = \left[ x^4 - \frac{5}{3}x^3 + 3x^2 + 9x \right]_{1}^{2}$$

Applying limits:

$$I = \left( 2^4 - \frac{5(2)^3}{3} + 3(2)^2 + 9(2) \right) - \left( 1^4 - \frac{5(1)^3}{3} + 3(1)^2 + 9(1) \right)$$ $$I = \left( 16 - \frac{40}{3} + 12 + 18 \right) - \left( 1 - \frac{5}{3} + 3 + 9 \right)$$ $$I = \left( 46 - \frac{40}{3} \right) - \left( 13 - \frac{5}{3} \right)$$ $$I = 46 - 13 - \frac{40}{3} + \frac{5}{3}$$ $$I = 33 - \frac{35}{3} = \frac{99 - 35}{3} = \frac{64}{3}$$

Answer: \(\frac{64}{3}\)

Question 4. Evaluate \(\int_{0}^{\pi/4} \sin 2x dx\)

Let \(I = \int_{0}^{\pi/4} \sin 2x dx\)

Integration of \(\sin ax\) is \(-\frac{\cos ax}{a}\).

$$I = \left[ -\frac{\cos 2x}{2} \right]_{0}^{\pi/4}$$ $$I = -\frac{1}{2} \left[ \cos 2(\frac{\pi}{4}) - \cos 2(0) \right]$$ $$I = -\frac{1}{2} \left[ \cos \frac{\pi}{2} - \cos 0 \right]$$ $$I = -\frac{1}{2} (0 - 1) = \frac{1}{2}$$

Answer: \(\frac{1}{2}\)

Question 5. Evaluate \(\int_{0}^{\pi/2} \cos 2x dx\)

Let \(I = \int_{0}^{\pi/2} \cos 2x dx\)

Integration of \(\cos ax\) is \(\frac{\sin ax}{a}\).

$$I = \left[ \frac{\sin 2x}{2} \right]_{0}^{\pi/2}$$ $$I = \frac{1}{2} \left[ \sin 2(\frac{\pi}{2}) - \sin 2(0) \right]$$ $$I = \frac{1}{2} (\sin \pi - \sin 0)$$ $$I = \frac{1}{2} (0 - 0) = 0$$

Answer: 0

Question 6. Evaluate \(\int_{4}^{5} e^x dx\)

Let \(I = \int_{4}^{5} e^x dx\)

$$I = \left[ e^x \right]_{4}^{5}$$ $$I = e^5 - e^4$$ $$I = e^4(e - 1)$$

Answer: \(e^4(e - 1)\)

Question 7. Evaluate \(\int_{0}^{\pi/4} \tan x dx\)

Let \(I = \int_{0}^{\pi/4} \tan x dx\)

We know \(\int \tan x dx = -\log|\cos x|\) or \(\log|\sec x|\).

$$I = \left[ -\log|\cos x| \right]_{0}^{\pi/4}$$ $$I = - [\log(\cos \frac{\pi}{4}) - \log(\cos 0)]$$ $$I = - [\log(\frac{1}{\sqrt{2}}) - \log(1)]$$ $$I = - [\log(2^{-1/2}) - 0]$$ $$I = - [-\frac{1}{2}\log 2] = \frac{1}{2}\log 2$$

Answer: \(\frac{1}{2}\log 2\)

Question 8. Evaluate \(\int_{\pi/6}^{\pi/4} \text{cosec } x dx\)

Let \(I = \int_{\pi/6}^{\pi/4} \text{cosec } x dx\)

Formula: \(\int \text{cosec } x dx = \log|\text{cosec } x - \cot x|\)

$$I = \left[ \log|\text{cosec } x - \cot x| \right]_{\pi/6}^{\pi/4}$$ $$I = \log|\text{cosec } \frac{\pi}{4} - \cot \frac{\pi}{4}| - \log|\text{cosec } \frac{\pi}{6} - \cot \frac{\pi}{6}|$$ $$I = \log|\sqrt{2} - 1| - \log|2 - \sqrt{3}|$$ $$I = \log \left( \frac{\sqrt{2} - 1}{2 - \sqrt{3}} \right)$$

Question 9. Evaluate \(\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}\)

Formula: \(\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1} x\)

$$I = \left[ \sin^{-1} x \right]_{0}^{1}$$ $$I = \sin^{-1}(1) - \sin^{-1}(0)$$ $$I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$

Answer: \(\frac{\pi}{2}\)

Question 10. Evaluate \(\int_{0}^{1} \frac{dx}{1+x^2}\)

Formula: \(\int \frac{dx}{1+x^2} = \tan^{-1} x\)

$$I = \left[ \tan^{-1} x \right]_{0}^{1}$$ $$I = \tan^{-1}(1) - \tan^{-1}(0)$$ $$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$

Answer: \(\frac{\pi}{4}\)

Question 11. Evaluate \(\int_{2}^{3} \frac{dx}{x^2 - 1}\)

Formula: \(\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right|\). Here \(a=1\).

$$I = \frac{1}{2} \left[ \log \left| \frac{x-1}{x+1} \right| \right]_{2}^{3}$$ $$I = \frac{1}{2} \left( \log \left| \frac{3-1}{3+1} \right| - \log \left| \frac{2-1}{2+1} \right| \right)$$ $$I = \frac{1}{2} \left( \log \frac{2}{4} - \log \frac{1}{3} \right)$$ $$I = \frac{1}{2} \left( \log \frac{1}{2} - \log \frac{1}{3} \right)$$ $$I = \frac{1}{2} \log \left( \frac{1/2}{1/3} \right) = \frac{1}{2} \log \left( \frac{3}{2} \right)$$

Answer: \(\frac{1}{2} \log \frac{3}{2}\)

Question 12. Evaluate \(\int_{0}^{\pi/2} \cos^2 x dx\)

Use identity: \(\cos^2 x = \frac{1 + \cos 2x}{2}\)

$$I = \int_{0}^{\pi/2} \frac{1 + \cos 2x}{2} dx$$ $$I = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_{0}^{\pi/2}$$ $$I = \frac{1}{2} \left[ (\frac{\pi}{2} + \frac{\sin \pi}{2}) - (0 + \frac{\sin 0}{2}) \right]$$ $$I = \frac{1}{2} \left( \frac{\pi}{2} + 0 - 0 \right) = \frac{\pi}{4}$$

Answer: \(\frac{\pi}{4}\)

Question 13. Evaluate \(\int_{2}^{3} \frac{x dx}{x^2+1}\)

Let \(x^2 + 1 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2}\)

Change limits: When \(x=2, t=5\); When \(x=3, t=10\).

$$I = \int_{5}^{10} \frac{1}{t} \frac{dt}{2} = \frac{1}{2} [\log t]_{5}^{10}$$ $$I = \frac{1}{2} (\log 10 - \log 5) = \frac{1}{2} \log(\frac{10}{5})$$ $$I = \frac{1}{2} \log 2$$

Answer: \(\frac{1}{2} \log 2\)

Question 14. Evaluate \(\int_{0}^{1} \frac{2x+3}{5x^2+1} dx\)

Split the integral: \(I = \int_{0}^{1} \frac{2x}{5x^2+1} dx + \int_{0}^{1} \frac{3}{5x^2+1} dx\)

Part 1: Multiply numerator and denominator by 5 to get derivative of denominator.

$$I_1 = \frac{1}{5} \int_{0}^{1} \frac{10x}{5x^2+1} dx = \frac{1}{5} [\log(5x^2+1)]_{0}^{1}$$ $$I_1 = \frac{1}{5} (\log 6 - \log 1) = \frac{1}{5} \log 6$$

Part 2: \(I_2 = 3 \int_{0}^{1} \frac{1}{5(x^2 + 1/5)} dx = \frac{3}{5} \int_{0}^{1} \frac{dx}{x^2 + (1/\sqrt{5})^2}\)

$$I_2 = \frac{3}{5} \cdot \frac{1}{1/\sqrt{5}} [\tan^{-1}(\frac{x}{1/\sqrt{5}})]_{0}^{1}$$ $$I_2 = \frac{3\sqrt{5}}{5} [\tan^{-1}(\sqrt{5}x)]_{0}^{1} = \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$$

Total Answer: \(\frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1} \sqrt{5}\)

Question 15. Evaluate \(\int_{0}^{1} x e^{x^2} dx\)

Let \(x^2 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2}\)

Limits: When \(x=0, t=0\); When \(x=1, t=1\).

$$I = \int_{0}^{1} e^t \frac{dt}{2} = \frac{1}{2} [e^t]_{0}^{1}$$ $$I = \frac{1}{2} (e^1 - e^0) = \frac{1}{2} (e - 1)$$

Answer: \(\frac{1}{2} (e - 1)\)

Question 16. Evaluate \(\int_{1}^{2} \frac{5x^2}{x^2+4x+3} dx\)

Divide \(5x^2\) by \(x^2+4x+3\). Quotient is 5, Remainder is \(-(20x+15)\).

$$I = \int_{1}^{2} \left( 5 - \frac{20x+15}{x^2+4x+3} \right) dx$$ $$I = [5x]_{1}^{2} - \int_{1}^{2} \frac{20x+15}{x^2+4x+3} dx$$ $$I = 5(2-1) - I_1 = 5 - I_1$$

Solving \(I_1\): Let \(20x+15 = A(2x+4) + B\). \(20x+15 = 20x + 40 - 25\).

$$I_1 = \int \frac{10(2x+4) - 25}{x^2+4x+3} dx$$ $$I_1 = 10 \int \frac{2x+4}{x^2+4x+3} dx - 25 \int \frac{dx}{x^2+4x+3}$$ $$I_1 = 10 \log|x^2+4x+3| - 25 \int \frac{dx}{(x+2)^2 - 1^2}$$ $$I_1 = 10 \log|x^2+4x+3| - 25 \cdot \frac{1}{2} \log|\frac{x+2-1}{x+2+1}|$$

Applying limits 1 to 2 for \(I_1\):

$$I_1 = \left[ 10 \log|x^2+4x+3| - \frac{25}{2} \log|\frac{x+1}{x+3}| \right]_{1}^{2}$$ $$I_1 = 10(\log 15 - \log 8) - \frac{25}{2}(\log \frac{3}{5} - \log \frac{2}{4})$$ $$I_1 = 10 \log 15 - 10 \log 8 - \frac{25}{2} \log 3 + \frac{25}{2} \log 5 + \frac{25}{2} \log \frac{1}{2}$$ Simplifying this expression results in \(I_1 = 10 \log 5 + 10 \log 3 - 30 \log 2 - \frac{25}{2} \log 3 + \frac{25}{2} \log 5 - \frac{25}{2} \log 2\) Alternatively, substituting back into \(I = 5 - I_1\).

Final simplified form: \(5 - [ 10 \log(\frac{15}{8}) - \frac{25}{2} \log(\frac{6}{5}) ]\)

Question 17. Evaluate \(\int_{0}^{\pi/4} (2\sec^2 x + x^3 + 2) dx\)

$$I = \left[ 2 \tan x + \frac{x^4}{4} + 2x \right]_{0}^{\pi/4}$$ $$I = \left( 2 \tan \frac{\pi}{4} + \frac{(\pi/4)^4}{4} + 2(\frac{\pi}{4}) \right) - (0)$$ $$I = 2(1) + \frac{\pi^4}{1024} + \frac{\pi}{2}$$ $$I = 2 + \frac{\pi}{2} + \frac{\pi^4}{1024}$$

Question 18. Evaluate \(\int_{0}^{\pi} (\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}) dx\)

Use formula: \(\cos^2 A - \sin^2 A = \cos 2A\).

$$\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos x$$

$$I = \int_{0}^{\pi} (-\cos x) dx = -[\sin x]_{0}^{\pi}$$ $$I = -(\sin \pi - \sin 0) = -(0 - 0) = 0$$

Answer: 0

Question 19. Evaluate \(\int_{0}^{2} \frac{6x+3}{x^2+4} dx\)

Split the integral:

$$I = \int_{0}^{2} \frac{6x}{x^2+4} dx + \int_{0}^{2} \frac{3}{x^2+4} dx$$ $$I = 3 \int_{0}^{2} \frac{2x}{x^2+4} dx + 3 \int_{0}^{2} \frac{dx}{x^2+2^2}$$

$$I = 3 [\log(x^2+4)]_{0}^{2} + 3 \cdot \frac{1}{2} [\tan^{-1} \frac{x}{2}]_{0}^{2}$$ $$I = 3 (\log 8 - \log 4) + \frac{3}{2} (\tan^{-1} 1 - \tan^{-1} 0)$$ $$I = 3 \log 2 + \frac{3}{2} (\frac{\pi}{4}) = 3 \log 2 + \frac{3\pi}{8}$$

Answer: \(3 \log 2 + \frac{3\pi}{8}\)

Question 20. Evaluate \(\int_{0}^{1} (x e^x + \sin \frac{\pi x}{4}) dx\)

Part 1: \(\int x e^x dx\). Using integration by parts: \(x e^x - \int e^x dx = x e^x - e^x\).

$$[x e^x - e^x]_{0}^{1} = (e - e) - (0 - 1) = 1$$

Part 2: \(\int \sin \frac{\pi x}{4} dx = \frac{-\cos(\pi x / 4)}{\pi / 4} = -\frac{4}{\pi} \cos \frac{\pi x}{4}\).

$$\left[ -\frac{4}{\pi} \cos \frac{\pi x}{4} \right]_{0}^{1} = -\frac{4}{\pi} (\cos \frac{\pi}{4} - \cos 0)$$ $$= -\frac{4}{\pi} (\frac{1}{\sqrt{2}} - 1) = \frac{4}{\pi} - \frac{2\sqrt{2}}{\pi}$$

Total \(I = 1 + \frac{4 - 2\sqrt{2}}{\pi}\)

Question 21. \(\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}\) equals

$$I = [\tan^{-1} x]_{1}^{\sqrt{3}}$$ $$I = \tan^{-1} \sqrt{3} - \tan^{-1} 1$$ $$I = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$$

A. \(\pi/3\)

B. \(2\pi/3\)

C. \(\pi/6\)

D. \(\pi/12\)

Correct Option: D

Question 22. \(\int_{0}^{2/3} \frac{dx}{4+9x^2}\) equals

$$I = \int_{0}^{2/3} \frac{dx}{(2)^2 + (3x)^2}$$ Using \(\int \frac{dx}{a^2 + (bx)^2} = \frac{1}{b} \cdot \frac{1}{a} \tan^{-1} \frac{bx}{a}\)

$$I = \frac{1}{3} \cdot \frac{1}{2} [\tan^{-1} \frac{3x}{2}]_{0}^{2/3}$$ $$I = \frac{1}{6} [\tan^{-1}(\frac{3}{2} \cdot \frac{2}{3}) - \tan^{-1} 0]$$ $$I = \frac{1}{6} (\tan^{-1} 1 - 0) = \frac{1}{6} \cdot \frac{\pi}{4} = \frac{\pi}{24}$$

A. \(\pi/6\)

B. \(\pi/12\)

C. \(\pi/24\)

D. \(\pi/4\)

Correct Option: C

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.6

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• Ex. 7.1


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• Ex. 7.2


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• Ex. 7.4


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• Ex. 7.8


• Ex. 7.9


• Ex. 7.10


• Ex. 7.11


• Miscellaneous Exercise

Integrals Class 12th Mathematics Part II CBSE Solution - Exercise 7.6

This exercise deals with Integration by Parts. The formula used is: $$ \int f(x)g(x)dx = f(x)\int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx $$ The choice of the first function \(f(x)\) and second function \(g(x)\) is usually based on the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential).

Question 1: Integrate the function \( x \sin x \)

Let \( I = \int x \sin x \, dx \).

Using integration by parts, let \( u = x \) (Algebraic) and \( v = \sin x \) (Trigonometric).

$$ I = x \int \sin x \, dx - \int \left( \frac{d}{dx}(x) \int \sin x \, dx \right) dx $$ $$ I = x(-\cos x) - \int 1 \cdot (-\cos x) \, dx $$ $$ I = -x \cos x + \int \cos x \, dx $$ $$ I = -x \cos x + \sin x + C $$

Question 2: Integrate the function \( x \sin 3x \)

Let \( I = \int x \sin 3x \, dx \).

Using integration by parts, take \( x \) as the first function and \( \sin 3x \) as the second function.

$$ I = x \int \sin 3x \, dx - \int \left( \frac{d}{dx}(x) \int \sin 3x \, dx \right) dx $$ $$ I = x\left(\frac{-\cos 3x}{3}\right) - \int 1 \cdot \left(\frac{-\cos 3x}{3}\right) dx $$ $$ I = -\frac{x \cos 3x}{3} + \frac{1}{3} \int \cos 3x \, dx $$ $$ I = -\frac{x \cos 3x}{3} + \frac{1}{9} \sin 3x + C $$

Question 3: Integrate the function \( x^2 e^x \)

Let \( I = \int x^2 e^x \, dx \).

Applying integration by parts with \( u = x^2 \) and \( v = e^x \):

$$ I = x^2 \int e^x \, dx - \int \left( \frac{d}{dx}(x^2) \int e^x \, dx \right) dx $$ $$ I = x^2 e^x - \int 2x e^x \, dx $$ $$ I = x^2 e^x - 2 \int x e^x \, dx $$

Applying integration by parts again for \( \int x e^x \, dx \):

$$ I = x^2 e^x - 2 \left[ x \int e^x \, dx - \int \left( \frac{d}{dx}(x) \int e^x \, dx \right) dx \right] $$ $$ I = x^2 e^x - 2 \left[ x e^x - \int e^x \, dx \right] $$ $$ I = x^2 e^x - 2(x e^x - e^x) + C $$ $$ I = e^x (x^2 - 2x + 2) + C $$

Question 4: Integrate the function \( x \log x \)

Let \( I = \int x \log x \, dx \).

Using ILATE, take Logarithmic (\( \log x \)) as first function and Algebraic (\( x \)) as second function.

$$ I = \log x \int x \, dx - \int \left( \frac{d}{dx}(\log x) \int x \, dx \right) dx $$ $$ I = \log x \left(\frac{x^2}{2}\right) - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2 \log x}{2} - \frac{1}{2} \int x \, dx $$ $$ I = \frac{x^2 \log x}{2} - \frac{x^2}{4} + C $$

Question 5: Integrate the function \( x \log 2x \)

Let \( I = \int x \log 2x \, dx \).

$$ I = \log 2x \int x \, dx - \int \left( \frac{d}{dx}(\log 2x) \int x \, dx \right) dx $$

Note: \( \frac{d}{dx}(\log 2x) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \).

$$ I = \frac{x^2}{2} \log 2x - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2 \log 2x}{2} - \frac{1}{2} \int x \, dx $$ $$ I = \frac{x^2 \log 2x}{2} - \frac{x^2}{4} + C $$

Question 6: Integrate the function \( x^2 \log x \)

Let \( I = \int x^2 \log x \, dx \).

$$ I = \log x \int x^2 \, dx - \int \left( \frac{d}{dx}(\log x) \int x^2 \, dx \right) dx $$ $$ I = \log x \left(\frac{x^3}{3}\right) - \int \frac{1}{x} \cdot \frac{x^3}{3} \, dx $$ $$ I = \frac{x^3 \log x}{3} - \frac{1}{3} \int x^2 \, dx $$ $$ I = \frac{x^3 \log x}{3} - \frac{x^3}{9} + C $$

Question 7: Integrate the function \( x \sin^{-1} x \)

Let \( I = \int x \sin^{-1} x \, dx \).

Take \( \sin^{-1} x \) as first function.

$$ I = \sin^{-1} x \left(\frac{x^2}{2}\right) - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \int \frac{-x^2}{\sqrt{1-x^2}} \, dx $$ $$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \int \frac{1-x^2-1}{\sqrt{1-x^2}} \, dx $$ $$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \int \left( \sqrt{1-x^2} - \frac{1}{\sqrt{1-x^2}} \right) dx $$

Using standard integrals \( \int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \).

$$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x - \sin^{-1} x \right] + C $$ $$ I = \frac{x^2 \sin^{-1} x}{2} + \frac{x}{4}\sqrt{1-x^2} - \frac{1}{4}\sin^{-1} x + C $$ $$ I = \frac{1}{4}(2x^2-1)\sin^{-1} x + \frac{x}{4}\sqrt{1-x^2} + C $$

Question 8: Integrate the function \( x \tan^{-1} x \)

Let \( I = \int x \tan^{-1} x \, dx \).

$$ I = \tan^{-1} x \left(\frac{x^2}{2}\right) - \int \frac{1}{1+x^2} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx $$ $$ I = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \frac{x^2+1-1}{1+x^2} \, dx $$ $$ I = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \left( 1 - \frac{1}{1+x^2} \right) dx $$ $$ I = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \left( x - \tan^{-1} x \right) + C $$

Question 9: Integrate the function \( x \cos^{-1} x \)

Let \( I = \int x \cos^{-1} x \, dx \).

Similar to Q7, but with \( \cos^{-1} x \).

$$ I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{-\sqrt{1-x^2}} \, dx $$ $$ I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \int \frac{1-x^2-1}{\sqrt{1-x^2}} \, dx $$ $$ I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \left[ \int \sqrt{1-x^2} \, dx - \int \frac{1}{\sqrt{1-x^2}} \, dx \right] $$ $$ I = \frac{x^2 \cos^{-1} x}{2} - \frac{1}{2} \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x - \sin^{-1} x \right] + C $$

Since \( \sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x \), we can absorb constants.

Final simplified form:

$$ I = \frac{2x^2-1}{4}\cos^{-1} x - \frac{x}{4}\sqrt{1-x^2} + C $$

Question 10: Integrate the function \( (\sin^{-1} x)^2 \)

Let \( I = \int (\sin^{-1} x)^2 \cdot 1 \, dx \).

$$ I = (\sin^{-1} x)^2 (x) - \int 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} \cdot x \, dx $$

For the integral part, put \( \sin^{-1} x = t \Rightarrow \frac{dx}{\sqrt{1-x^2}} = dt \) and \( x = \sin t \).

$$ \int 2t \sin t \, dt = 2 \left( t(-\cos t) - \int 1(-\cos t)dt \right) = -2t \cos t + 2 \sin t $$

Substitute back \( t = \sin^{-1} x \), \( \cos t = \sqrt{1-x^2} \).

$$ I = x(\sin^{-1} x)^2 - [-2\sqrt{1-x^2}\sin^{-1} x + 2x] + C $$ $$ I = x(\sin^{-1} x)^2 + 2\sqrt{1-x^2}\sin^{-1} x - 2x + C $$

Question 11: Integrate the function \( \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \)

Let \( I = \int \cos^{-1} x \cdot \frac{x}{\sqrt{1-x^2}} \, dx \).

Put \( \cos^{-1} x = t \Rightarrow \frac{-1}{\sqrt{1-x^2}} dx = dt \Rightarrow \frac{dx}{\sqrt{1-x^2}} = -dt \). Also \( x = \cos t \).

$$ I = \int t \cdot \cos t \cdot (-dt) = -\int t \cos t \, dt $$

Integrating by parts:

$$ I = - [ t \sin t - \int \sin t \, dt ] $$ $$ I = - [ t \sin t + \cos t ] + C $$

Substituting back \( t = \cos^{-1} x \), \( \sin t = \sqrt{1-x^2} \):

$$ I = - [\sqrt{1-x^2} \cos^{-1} x + x] + C $$

Question 12: Integrate the function \( x \sec^2 x \)

Let \( I = \int x \sec^2 x \, dx \).

$$ I = x \int \sec^2 x \, dx - \int \left( \frac{d}{dx}(x) \int \sec^2 x \, dx \right) dx $$ $$ I = x \tan x - \int \tan x \, dx $$ $$ I = x \tan x - \log|\sec x| + C $$

Or \( I = x \tan x + \log|\cos x| + C \)

Question 13: Integrate the function \( \tan^{-1} x \)

Let \( I = \int 1 \cdot \tan^{-1} x \, dx \).

$$ I = \tan^{-1} x (x) - \int \frac{1}{1+x^2} \cdot x \, dx $$ $$ I = x \tan^{-1} x - \frac{1}{2} \int \frac{2x}{1+x^2} \, dx $$ $$ I = x \tan^{-1} x - \frac{1}{2} \log|1+x^2| + C $$

Question 14: Integrate the function \( x (\log x)^2 \)

Let \( I = \int (\log x)^2 \cdot x \, dx \).

$$ I = (\log x)^2 \frac{x^2}{2} - \int 2 \log x \cdot \frac{1}{x} \cdot \frac{x^2}{2} \, dx $$ $$ I = \frac{x^2}{2} (\log x)^2 - \int x \log x \, dx $$

Using result from Q4 for \( \int x \log x \, dx \):

$$ I = \frac{x^2}{2} (\log x)^2 - \left[ \frac{x^2}{2} \log x - \frac{x^2}{4} \right] + C $$ $$ I = \frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C $$

Question 15: Integrate the function \( (x^2+1) \log x \)

Let \( I = \int \log x (x^2+1) \, dx \).

$$ I = \log x \left( \frac{x^3}{3} + x \right) - \int \frac{1}{x} \left( \frac{x^3}{3} + x \right) dx $$ $$ I = \left( \frac{x^3}{3} + x \right) \log x - \int \left( \frac{x^2}{3} + 1 \right) dx $$ $$ I = \left( \frac{x^3}{3} + x \right) \log x - \frac{x^3}{9} - x + C $$

Question 16: Integrate the function \( e^x (\sin x + \cos x) \)

We use the property: \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \).

Here, let \( f(x) = \sin x \), then \( f'(x) = \cos x \).

Thus, \( I = e^x \sin x + C \).

Question 17: Integrate the function \( \frac{x e^x}{(1+x)^2} \)

Rewrite integrand:

$$ I = \int e^x \left[ \frac{1+x-1}{(1+x)^2} \right] dx $$ $$ I = \int e^x \left[ \frac{1}{1+x} - \frac{1}{(1+x)^2} \right] dx $$

Let \( f(x) = \frac{1}{1+x} \), then \( f'(x) = \frac{-1}{(1+x)^2} \).

Using the property, \( I = \frac{e^x}{1+x} + C \).

Question 18: Integrate the function \( e^x \left(\frac{1+\sin x}{1+\cos x}\right) \)

Using half-angle formulas:

$$ \frac{1+\sin x}{1+\cos x} = \frac{1 + 2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} $$ $$ = \frac{1}{2\cos^2(x/2)} + \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} $$ $$ = \frac{1}{2}\sec^2(x/2) + \tan(x/2) $$

Integral becomes \( \int e^x [ \tan(x/2) + \frac{1}{2}\sec^2(x/2) ] dx \).

Let \( f(x) = \tan(x/2) \), then \( f'(x) = \frac{1}{2}\sec^2(x/2) \).

Thus, \( I = e^x \tan(x/2) + C \).

Question 19: Integrate the function \( e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) \)

Let \( f(x) = \frac{1}{x} \), then \( f'(x) = -\frac{1}{x^2} \).

Using the standard property, \( I = \frac{e^x}{x} + C \).

Question 20: Integrate the function \( \frac{(x-3)e^x}{(x-1)^3} \)

Rewrite numerator \( x-3 \) as \( (x-1) - 2 \).

$$ I = \int e^x \left[ \frac{(x-1)-2}{(x-1)^3} \right] dx $$ $$ I = \int e^x \left[ \frac{1}{(x-1)^2} + \frac{-2}{(x-1)^3} \right] dx $$

Let \( f(x) = (x-1)^{-2} \), then \( f'(x) = -2(x-1)^{-3} \).

Thus, \( I = \frac{e^x}{(x-1)^2} + C \).

Question 21: Integrate the function \( e^{2x} \sin x \)

Let \( I = \int e^{2x} \sin x \, dx \).

By parts: \( u=\sin x, v=e^{2x} \).

$$ I = \sin x \frac{e^{2x}}{2} - \int \cos x \frac{e^{2x}}{2} \, dx $$

Apply parts again on integral:

$$ \int e^{2x} \cos x \, dx = \cos x \frac{e^{2x}}{2} - \int (-\sin x) \frac{e^{2x}}{2} \, dx $$

Substitute back:

$$ I = \frac{e^{2x}\sin x}{2} - \frac{1}{2} \left[ \frac{e^{2x}\cos x}{2} + \frac{1}{2} I \right] $$ $$ I = \frac{e^{2x}\sin x}{2} - \frac{e^{2x}\cos x}{4} - \frac{1}{4} I $$ $$ \frac{5}{4} I = \frac{e^{2x}}{4} (2\sin x - \cos x) $$ $$ I = \frac{e^{2x}}{5} (2\sin x - \cos x) + C $$

Question 22: Integrate the function \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)

Put \( x = \tan \theta \implies dx = \sec^2 \theta \, d\theta \).

$$ \sin^{-1}(\sin 2\theta) = 2\theta $$ $$ I = \int 2\theta \sec^2 \theta \, d\theta $$

By parts: \( 2 [ \theta \tan \theta - \int \tan \theta \, d\theta ] \).

$$ I = 2\theta \tan \theta - 2 \log|\sec \theta| + C $$

Substitute \( \theta = \tan^{-1} x \). Note \( \log|\sec \theta| = \frac{1}{2}\log(1+\tan^2 \theta) = \frac{1}{2}\log(1+x^2) \).

$$ I = 2x \tan^{-1} x - \log(1+x^2) + C $$

Question 23: \( \int x^2 e^{x^3} \, dx \) equals

Let \( x^3 = t \implies 3x^2 dx = dt \implies x^2 dx = \frac{dt}{3} \).

$$ I = \int e^t \frac{dt}{3} = \frac{1}{3} e^t + C = \frac{1}{3} e^{x^3} + C $$

A. \( \frac{1}{3} e^{x^3} + C \)

Question 24: \( \int e^x \sec x (1 + \tan x) \, dx \) equals

$$ I = \int e^x (\sec x + \sec x \tan x) \, dx $$

Let \( f(x) = \sec x \), then \( f'(x) = \sec x \tan x \).

By standard property, \( I = e^x \sec x + C \).

B. \( e^x \sec x + C \)