JEE Mathematics Indefinite Integration MCQs Set 11

Answer: (a) \( \log{\left(\sqrt{x + 1} \right)} - \frac{\log{\left(\sqrt{x^{2} + 1} \right)}}{2} + \frac{\tan^{-1}{\left(x \right)}}{2}+c \)

Explanation:
1. \( 1+x+x^2+x^3=(1+x)(1+x^2) \) and \( \frac{1}{(1+x)(1+x^2)}=\frac{1/2}{1+x}+\frac{\frac12(1-x)}{1+x^2} \).
2. \( \int=\frac12\log(1+x)-\frac14\log(1+x^2)+\frac12\tan^{-1}x \), which is \( \log\sqrt{1+x}-\frac12\log\sqrt{1+x^2}+\frac12\tan^{-1}x+c \).

Teacher's Note:
1. Factorise by grouping.
2. Split with partial fractions with a quadratic factor.

Answer: (b) \( x - \log{\left(x \right)} + \log{\left(\sqrt{x^{2} + 1} \right)} - \tan^{-1}{\left(x \right)}+c \)

Explanation:
1. \( \frac{x^3-1}{x(x^2+1)}=1-\frac{x+1}{x(x^2+1)}=1-\frac1x+\frac{x-1}{x^2+1} \).
2. \( \int=x-\log x+\frac12\log(x^2+1)-\tan^{-1}x+c \), and \( \frac12\log(x^2+1)=\log\sqrt{x^2+1} \).

Teacher's Note:
1. Divide first and then use partial fractions.
2. The term (x - 1) over x squared + 1 gives a log and an arctangent.

Answer: (c) \( -x-6\log|1-x|-\dfrac{12}{1-x}+\dfrac{4}{(1-x)^2}+c \)

Explanation:
1. Put \( u=1-x \): \( \frac{(2-u)^3}{u^3}=\frac{8}{u^3}-\frac{12}{u^2}+\frac6u-1 \) and \( dx=-du \).
2. \( -\int\left[\frac8{u^3}-\frac{12}{u^2}+\frac6u-1\right]du=\frac{4}{u^2}-\frac{12}{u}-6\log|u|+u \).
3. Substituting \( u=1-x \): \( -x-6\log|1-x|-\frac{12}{1-x}+\frac{4}{(1-x)^2}+c \) (the constant 1 is absorbed in c).

Teacher's Note:
1. Substitute u = 1 - x so that the numerator is 2 - u.
2. Remember dx = - du.

Answer: (c) \( A=-\dfrac32,\ B=\dfrac{35}{36},\ C=\text{an arbitrary constant} \)

Explanation:
1. Multiply numerator and denominator by \( e^x \): \( \frac{4e^{2x}+6}{9e^{2x}-4}=\frac49+\frac{70/9}{9e^{2x}-4} \).
2. \( \int\frac{dx}{9e^{2x}-4}=-\frac x4+\frac18\log(9e^{2x}-4) \).
3. So the integral is \( \frac{4x}{9}-\frac{35x}{18}+\frac{35}{36}\log(9e^{2x}-4)=-\frac32x+\frac{35}{36}\log(9e^{2x}-4)+C \).

Teacher's Note:
1. Multiply by e to the x to get a function of e to the 2x.
2. Write 1 over (9u - 4) as minus 1/4 plus 9u over 4(9u - 4).

Answer: (a) \( \frac{\log{\left(\frac{x^{2} - 1}{x^{2} + 1} \right)}}{4}+c \)

Explanation:
1. Put \( u=x^2 \), \( x\,dx=\frac{du}2 \): \( \frac12\int\frac{du}{u^2-1} \).
2. \( \frac12\cdot\frac12\log\frac{u-1}{u+1}=\frac14\log\frac{x^2-1}{x^2+1}+c \).

Teacher's Note:
1. Substitute u = x squared.
2. Use 1 over u squared minus 1 as a standard form.

Answer: (a) \( \frac{\log{\left(1 - \cos{\left(x \right)} \right)}}{6} + \frac{\log{\left(\cos{\left(x \right)} + 1 \right)}}{2} - \frac{2 \log{\left(2 \cos{\left(x \right)} + 1 \right)}}{3}+c \)

Explanation:
1. \( \sin x+\sin2x=\sin x(1+2\cos x) \), so \( \frac{1}{\sin x(1+2\cos x)}=\frac{\sin x}{(1-\cos^2x)(1+2\cos x)} \).
2. Put \( t=\cos x \): \( -\int\frac{dt}{(1-t)(1+t)(1+2t)} \) with \( \frac{1}{(1-t)(1+t)(1+2t)}=\frac{1/6}{1-t}-\frac{1/2}{1+t}+\frac{4/3}{1+2t} \).
3. The integral is \( \frac16\log(1-\cos x)+\frac12\log(1+\cos x)-\frac23\log(1+2\cos x)+c \).

Teacher's Note:
1. Multiply numerator and denominator by sin x.
2. Then substitute t = cos x and use partial fractions.

Answer: (c) \( \sin{\left(x \right)} - 6 \tan^{-1}{\left(\sin{\left(x \right)} \right)} - \frac{2}{\sin{\left(x \right)}}+c \)

Explanation:
1. Put \( t=\sin x \), \( \cos x\,dx=dt \): \( \int\frac{(1-t^2)(2-t^2)}{t^2(1+t^2)}dt \).
2. \( \frac{t^4-3t^2+2}{t^4+t^2}=1+\frac{2}{t^2}-\frac{6}{1+t^2} \).
3. The integral is \( t-\frac2t-6\tan^{-1}t=\sin x-2(\sin x)^{-1}-6\tan^{-1}(\sin x)+c \).

Teacher's Note:
1. Take out cos x and write the rest in terms of sin x.
2. Then divide the polynomial and split.

Answer: (a) \( \frac{\log{\left(\tan{\left(x \right)} + \sec{\left(x \right)} \right)}}{2} + \frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2}+c \)

Explanation:
1. By parts: \( I=\sec x\tan x-\int\sec x\tan^2x\,dx=\sec x\tan x-I+\int\sec x\,dx \).
2. So \( 2I=\sec x\tan x+\log(\sec x+\tan x) \) and \( I=\frac12\left[\sec x\tan x+\log(\sec x+\tan x)\right]+c \).

Teacher's Note:
1. The integral reappears on the right side, so solve for it.
2. Memorise this standard result.

Answer: (b) \( - \frac{3}{\sqrt[3]{\tan{\left(x \right)}}}+c \)

Explanation:
1. \( \sec^{2/3}x\operatorname{cosec}^{4/3}x=\frac{1}{\cos^{2/3}x\sin^{4/3}x}=\frac{\sec^2x}{\tan^{4/3}x} \).
2. Put \( t=\tan x \): \( \int t^{-4/3}dt=-3t^{-1/3}=-3(\tan x)^{-1/3}+c \).

Teacher's Note:
1. Divide numerator and denominator by cos squared x to form tan x.
2. Apply the power rule with exponent minus 4/3.

Answer: (b) \( - \frac{\sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}+c \)

Explanation:
1. \( \sin^4x\cos^3x=\sin^4x(1-\sin^2x)\cos x \).
2. Put \( t=\sin x \): \( \int(t^4-t^6)dt=\frac{\sin^5x}{5}-\frac{\sin^7x}{7}+c \).

Teacher's Note:
1. Keep one cos x as dt and convert the other cos squared.
2. Then integrate the polynomial.

Answer: (b) \( -\dfrac{3}{16} \)

Explanation:
1. \( \sin3x\cos5x=\frac12(\sin8x-\sin2x) \).
2. \( \int=\frac12\left(-\frac{\cos8x}{8}+\frac{\cos2x}{2}\right)+C \).
3. At \( x=0 \): \( \frac12\left(-\frac18+\frac12\right)+C=\frac{3}{16}+C=0 \), so \( C=-\frac3{16} \).

Teacher's Note:
1. Use the product to sum formula first.
2. Substitute x = 0 to find the constant.

Answer: (c) \( - \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left(6 x \right)}}{12}+c \)

Explanation:
1. \( \cos2x\sin4x=\frac12(\sin6x+\sin2x) \).
2. \( \int=\frac12\left(-\frac{\cos6x}{6}-\frac{\cos2x}{2}\right)=-\left(\frac{\cos2x}{4}+\frac{\cos6x}{12}\right)+c \).

Teacher's Note:
1. Use 2 sin A cos B as a sum of sines.
2. Do not forget the factor one half.

Answer: (d) \( x \)

Explanation:
1. \( \cot^4x=\cot^2x(\operatorname{cosec}^2x-1)=\cot^2x\operatorname{cosec}^2x-\operatorname{cosec}^2x+1 \).
2. So \( \int\cot^4x\,dx=-\frac{\cot^3x}{3}+\cot x+x+c \).
3. Adding the terms outside the integral, \( f(x)=-\frac{\cot^3x}{3}+\cot x+x+\frac13\cot^3x-\cot x+c=x+c \).
4. \( f\left(\frac\pi2\right)=\frac\pi2 \) gives \( c=0 \), so \( f(x)=x \).

Teacher's Note:
1. Reduce cot to the fourth with cosec squared.
2. Use the given value to fix the constant.

Answer: (b) \( \cosh^{-1}\left(\dfrac{\log x}{\sqrt3}\right)+c \)

Explanation:
1. Put \( t=\log x \), \( dt=\frac{dx}{x} \): \( \int\frac{dt}{\sqrt{t^2-3}} \).
2. \( \int\frac{dt}{\sqrt{t^2-a^2}}=\cosh^{-1}\frac ta \), so the result is \( \cosh^{-1}\left(\frac{\log x}{\sqrt3}\right)+c \).

Teacher's Note:
1. Substitute t = log x to remove x.
2. Use the inverse hyperbolic cosine form with a equal to root 3.

Answer: (d) \( x - \coth{\left(x \right)}+c \)

Explanation:
1. The integrand is \( \coth^2x=1+\operatorname{cosech}^2x \).
2. \( \int(1+\operatorname{cosech}^2x)dx=x-\coth x+c \).

Teacher's Note:
1. Recognise the ratio as coth squared.
2. The integral of cosech squared is minus coth.

Answer: (c) \( -2\cosh^{-1}\left\{\sqrt2\cos\left(\dfrac x2\right)\right\}+c \)

Explanation:
1. \( \sec x-1=\frac{1-\cos x}{\cos x}=\frac{2(1-u^2)}{2u^2-1} \) with \( u=\cos\frac x2 \), and \( dx=-\frac{2\,du}{\sqrt{1-u^2}} \).
2. The integral becomes \( -2\sqrt2\int\frac{du}{\sqrt{2u^2-1}}=-2\int\frac{du}{\sqrt{u^2-\frac12}}=-2\cosh^{-1}(\sqrt2u) \).
3. So the result is \( -2\cosh^{-1}\left\{\sqrt2\cos\frac x2\right\}+c \).

Teacher's Note:
1. Express cos x through cos of half the angle.
2. The final integral is the inverse hyperbolic cosine form.

Answer: (b) \( e^x\cosh x+c \)

Explanation:
1. \( \sinh x+\cosh x=e^x \), so the integrand is \( e^{2x} \).
2. \( \int e^{2x}dx=\frac{e^{2x}}{2}=e^x\cosh x-\frac12+\text{const.} \), since \( e^x\cosh x=\frac{e^{2x}+1}{2} \).
3. So the integral is \( e^x\cosh x+c \).

Teacher's Note:
1. Use sinh x + cosh x = e to the x.
2. Check the options by differentiating: the derivative of e to the x cosh x is e to the x (cosh x + sinh x).

Answer: (a) \( \frac{2 \log{\left(x^{\frac{7}{2}} + \sqrt{x^{7} + 1} \right)}}{7}+c \)

Explanation:
1. Put \( u=x^{7/2} \), \( du=\frac72x^{5/2}dx \): \( \frac27\int\frac{du}{\sqrt{1+u^2}} \).
2. \( \frac27\log\left(u+\sqrt{1+u^2}\right)=\frac27\log\left(x^{7/2}+\sqrt{x^7+1}\right)+c \).

Teacher's Note:
1. Substitute u equal to x to the 7/2 so that u squared is x to the 7.
2. Then use the standard logarithmic form.

Answer: (a) \( \frac{4 \left(1 - \frac{1}{x^{3}}\right)^{\frac{5}{4}}}{15}+c \)

Explanation:
1. \( (x^4-x)^{1/4}=x\left(1-\frac1{x^3}\right)^{1/4} \), so the integrand is \( x^{-4}\left(1-x^{-3}\right)^{1/4} \).
2. Put \( u=1-x^{-3} \), \( du=3x^{-4}dx \): \( \frac13\int u^{1/4}du=\frac13\cdot\frac45u^{5/4}=\frac4{15}\left(1-\frac1{x^3}\right)^{5/4}+c \).

Teacher's Note:
1. Take x to the fourth out of the bracket.
2. The remaining factor x to the minus 4 is the derivative of x to the minus 3 up to a constant.

Answer: (a) \( \dfrac13 \)

Explanation:
1. Put \( t=\sqrt{1-x^3} \): \( x^3=1-t^2 \) and \( \frac{dx}{x}=-\frac{2t\,dt}{3(1-t^2)} \).
2. The integral becomes \( -\frac23\int\frac{dt}{1-t^2}=-\frac13\log\left|\frac{1+t}{1-t}\right|=\frac13\log\left|\frac{t-1}{t+1}\right| \).
3. So \( a=\frac13 \).

Teacher's Note:
1. Substitute the square root to get a rational function.
2. Express dx over x through t.

Answer: (b) \( \log(1+\sqrt2)-\dfrac\pi4 \)

Explanation:
1. Put \( x=\tan\theta \): the integrand becomes \( \frac{\tan^2\theta}{1+\sec\theta}d\theta=\frac{\sec^2\theta-1}{1+\sec\theta}d\theta=(\sec\theta-1)d\theta \).
2. \( f=\log(\sec\theta+\tan\theta)-\theta+c \) and \( f(0)=0 \) gives \( c=0 \).
3. At \( x=1 \), \( \theta=\frac\pi4 \): \( f(1)=\log(\sqrt2+1)-\frac\pi4 \).

Teacher's Note:
1. The substitution x = tan theta simplifies the square root to sec theta.
2. Use the condition f(0) = 0 to fix the constant.

Answer: (b) \( f(x)=x^{3/2} \) and \( g(x)=\sin^{-1}x \)

Explanation:
1. Put \( u=x^{3/2} \), \( du=\frac32x^{1/2}dx \).
2. \( \frac23\int\frac{du}{\sqrt{1-u^2}}=\frac23\sin^{-1}u=\frac23\sin^{-1}\left(x^{3/2}\right)+c \), so \( f(x)=x^{3/2} \) and \( g(x)=\sin^{-1}x \).

Teacher's Note:
1. Choose u so that u squared is x cubed.
2. The result is then the standard arcsine form.

Answer: (d) \( \dfrac{7}{\alpha+\beta}\left(\dfrac{x-\beta}{x+\alpha}\right)^{1/7} \)

Explanation:
1. Write the integrand as \( \left(\frac{x-\beta}{x+\alpha}\right)^{-6/7}\frac{1}{(x+\alpha)^2} \).
2. Put \( u=\frac{x-\beta}{x+\alpha} \); then \( du=\frac{\alpha+\beta}{(x+\alpha)^2}dx \).
3. \( \frac{1}{\alpha+\beta}\int u^{-6/7}du=\frac{7}{\alpha+\beta}u^{1/7}=\frac{7}{\alpha+\beta}\left(\frac{x-\beta}{x+\alpha}\right)^{1/7}+c \).

Teacher's Note:
1. The exponents add up to 2, which suggests the quotient substitution.
2. The derivative of (x - beta) over (x + alpha) is (alpha + beta) over (x + alpha) squared.

Answer: (c) \( \dfrac52\left(\dfrac{x^2+1}{x^2+2}\right)^{1/5}+c \)

Explanation:
1. The integrand is \( x\left(\frac{x^2+1}{x^2+2}\right)^{-4/5}\frac{1}{(x^2+2)^2} \).
2. Put \( u=\frac{x^2+1}{x^2+2} \); then \( du=\frac{2x\,dx}{(x^2+2)^2} \).
3. \( \frac12\int u^{-4/5}du=\frac52u^{1/5}=\frac52\left(\frac{x^2+1}{x^2+2}\right)^{1/5}+c \).

Teacher's Note:
1. The exponents 4/5 and 6/5 add up to 2, which suggests the quotient substitution.
2. Differentiate (x squared + 1) over (x squared + 2) to confirm the numerator.