JEE Mathematics Indefinite Integration MCQs Set 05

Answer: (a) \( x^{x}+c \)

Explanation:
1. \( \frac{d}{dx}(x^x)=x^x(1+\log x) \) by logarithmic differentiation.
2. Hence the integral is \( x^x+c \).

Teacher's Note:
1. Recognise the derivative of x to the power x.
2. Logarithmic differentiation gives the factor 1 + log x.

Answer: (c) \( \frac{\left(x + \log{\left(x \right)}\right)^{3}}{3}+c \)

Explanation:
1. Put \( t=x+\log x \); then \( dt=\left(1+\frac1x\right)dx=\frac{x+1}{x}dx \).
2. \( \int t^2dt=\frac{t^3}{3}=\frac{(x+\log x)^3}{3}+c \).

Teacher's Note:
1. Look for a function whose derivative also appears in the integrand.
2. Here (x + 1)/x is the derivative of x + log x.

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

Explanation:
1. Divide numerator and denominator by \( x^2 \): the integrand is \( \frac{1+1/x^2}{x-1/x} \).
2. Put \( t=x-\frac1x \), so \( dt=\left(1+\frac{1}{x^2}\right)dx \).
3. \( \int\frac{dt}{t}=\log\left|x-\frac1x\right|=\log\frac{x^2-1}{x}+C \) for \( x>1 \).

Teacher's Note:
1. Dividing by x squared exposes the derivative of x minus 1/x in the numerator.
2. Partial fractions give the same result.

Answer: (b) \( - \frac{\sqrt[4]{x^{4} + 1}}{x}+c \)

Explanation:
1. Write the integrand as \( \frac{1}{x^5\left(1+x^{-4}\right)^{3/4}} \) and put \( t=1+x^{-4} \), so \( dt=-4x^{-5}dx \).
2. \( -\frac14\int t^{-3/4}dt=-t^{1/4}=-\frac{(x^4+1)^{1/4}}{x}+c \).

Teacher's Note:
1. Take x to the fourth out of the bracket to create a negative power of x.
2. Then substitute for the bracket.

Answer: (d) \( \frac{\sin{\left(x^{2} \right)}}{2}+c \)

Explanation:
1. Put \( t=x^2 \), \( x\,dx=\frac{dt}{2} \).
2. \( \frac12\int\cos t\,dt=\frac12\sin x^2+C \).

Teacher's Note:
1. The substitution t = x squared leaves a basic integral.
2. Do not confuse sin of x squared with sin squared x.

Answer: (a) \( \frac{\cot{\left(5 + 2 e^{- x} \right)}}{2}+c \)

Explanation:
1. Put \( t=2e^{-x}+5 \); then \( dt=-2e^{-x}dx \).
2. \( -\frac12\int\operatorname{cosec}^2t\,dt=\frac12\cot t+c=\frac12\cot(2e^{-x}+5)+c \).

Teacher's Note:
1. Two negative signs cancel: one from dt and one from the integral of cosec squared.
2. Divide by the factor 2 from the chain rule.

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

Explanation:
1. The derivative of \( x+\log\sec x \) is \( 1+\tan x \).
2. So the integral is \( \log(x+\log\sec x)+c \).

Teacher's Note:
1. Check whether the numerator is the derivative of the denominator.
2. Then the integral is the log of the denominator.

Answer: (b) \( \frac{\left(x^{2} + 2\right)^{\frac{3}{2}}}{3} - 2 \sqrt{x^{2} + 2}+c \)

Explanation:
1. Put \( t=x^2+2 \), so \( x^2=t-2 \) and \( x\,dx=\frac{dt}{2} \).
2. \( \frac12\int\frac{t-2}{\sqrt t}dt=\frac12\left(\frac23t^{3/2}-4t^{1/2}\right)=\frac13t^{3/2}-2t^{1/2} \).

Teacher's Note:
1. Split x cubed as x squared times x before substituting.
2. Replace x squared by t minus 2.

Answer: (c) \( \frac{\left(5 x^{4} + 3\right)^{\frac{3}{2}}}{30}+c \)

Explanation:
1. Put \( t=3+5x^4 \), \( dt=20x^3dx \).
2. \( \frac{1}{20}\int t^{1/2}dt=\frac{1}{20}\cdot\frac23t^{3/2}=\frac{1}{30}(3+5x^4)^{3/2}+c \).

Teacher's Note:
1. Multiply the factor 1/20 by 2/3 from the power rule.
2. The result is 1/30.

Answer: (c) \( \frac{\sin^{3}{\left(x \right)}}{3}+c \)

Explanation:
1. Put \( t=\sin x \), \( dt=\cos x\,dx \).
2. \( \int t^2dt=\frac{\sin^3x}{3}+c \).

Teacher's Note:
1. The derivative of sin x is present as a factor.
2. Then apply the power rule.

Answer: (c) \( \frac{\left(\log{\left(x \right)} + 1\right)^{3}}{3}+c \)

Explanation:
1. Put \( t=1+\log x \), \( dt=\frac{dx}{x} \).
2. \( \int t^2dt=\frac{(1+\log x)^3}{3}+c \).

Teacher's Note:
1. The factor 1/x is the derivative of log x.
2. Remember to divide by 3.

Answer: (b) \( \frac{1}{b \left(a + b \cos{\left(x \right)}\right)}+c \)

Explanation:
1. Put \( t=a+b\cos x \), so \( dt=-b\sin x\,dx \).
2. \( -\frac1b\int t^{-2}dt=\frac{1}{bt}=\frac{1}{b(a+b\cos x)}+c \).

Teacher's Note:
1. Two minus signs again cancel.
2. The exponent minus 2 integrates to minus 1 over t.

Answer: (a) \( - \frac{\cos^{3}{\left(a x + b \right)}}{3 a}+c \)

Explanation:
1. Put \( t=\cos(ax+b) \), so \( dt=-a\sin(ax+b)dx \).
2. \( -\frac1a\int t^2dt=-\frac{\cos^3(ax+b)}{3a}+c \).

Teacher's Note:
1. The factor a comes from the chain rule.
2. The sign comes from the derivative of cosine.

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

Explanation:
1. Put \( t=\tan x \), \( dt=\sec^2x\,dx=(1+t^2)dx \).
2. \( \int\frac{dt}{1-t^2}=\frac12\log\frac{1+t}{1-t}=\frac12\log\frac{1+\tan x}{1-\tan x}+c \).

Teacher's Note:
1. Write dx as dt over 1 + t squared so the numerator cancels.
2. Use the standard 1 over (1 - t squared) integral.

Answer: (b) \( 2a^{\sqrt x}\log_a e+c \)

Explanation:
1. Put \( t=\sqrt x \), \( dx=2t\,dt \); the integral becomes \( 2\int a^tdt=\frac{2a^t}{\log a} \).
2. Since \( \frac{1}{\log_ea}=\log_ae \), the result is \( 2a^{\sqrt x}\log_ae+c \).

Teacher's Note:
1. Substitute the root and remember the factor 2.
2. One over natural log of a equals log of e to the base a.

Answer: (d) \( - \frac{e^{- 3 t^{2}}}{6}+c \)

Explanation:
1. Write the integrand as \( t\,e^{-3t^2} \) and put \( u=-3t^2 \), so \( t\,dt=-\frac{du}{6} \).
2. \( -\frac16\int e^udu=-\frac16e^{-3t^2}+c \).

Teacher's Note:
1. Bring the exponential to the numerator with a negative power.
2. The derivative of the exponent is minus 6t.

Answer: (b) \( \frac{e^{x^{2}}}{2}+c \)

Explanation:
1. Put \( t=x^2 \), \( x\,dx=\frac{dt}{2} \).
2. \( \frac12\int e^tdt=\frac{e^{x^2}}{2}+C \).

Teacher's Note:
1. The derivative of x squared is 2x.
2. Divide by 2 after substituting.

Answer: (c) \( \left(2 \sqrt{x} - 2\right) e^{\sqrt{x}}+c \)

Explanation:
1. Put \( t=\sqrt x \), \( dx=2t\,dt \): \( 2\int te^tdt \).
2. By parts, \( \int te^tdt=(t-1)e^t \), so the result is \( 2(\sqrt x-1)e^{\sqrt x}+A \).

Teacher's Note:
1. Substitution followed by integration by parts.
2. Check by differentiating the answer.

Answer: (c) \( 2 e^{\sqrt{x}}+c \)

Explanation:
1. Put \( t=\sqrt x \), \( dt=\frac{dx}{2\sqrt x} \).
2. \( 2\int e^tdt=2e^{\sqrt x}+c \).

Teacher's Note:
1. The derivative of root x is 1 over 2 root x.
2. That gives the factor 2.

Answer: (a) \( \frac{a \log{\left(\frac{e^{x}}{b + c e^{x}} \right)}}{b}+c \)

Explanation:
1. Multiply numerator and denominator by \( e^{-x} \): \( \frac{ae^{-x}}{be^{-x}+c} \).
2. Put \( t=be^{-x}+c \), \( dt=-be^{-x}dx \): \( -\frac ab\log t=-\frac ab\log\frac{b+ce^x}{e^x}=\frac ab\log\frac{e^x}{b+ce^x}+k \).

Teacher's Note:
1. Multiply by e to the minus x to make the numerator a derivative.
2. Keep the factor a over b.

Answer: (a) \( - \frac{1}{2 e^{2 x} + 2}+c \)

Explanation:
1. \( \frac{1}{(e^x+e^{-x})^2}=\frac{e^{2x}}{(e^{2x}+1)^2} \).
2. Put \( t=e^{2x}+1 \), \( dt=2e^{2x}dx \): \( \frac12\int t^{-2}dt=-\frac{1}{2(e^{2x}+1)}+c \).

Teacher's Note:
1. Multiply numerator and denominator by e to the 2x.
2. The new numerator is half the derivative of the base.

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

Explanation:
1. Write \( \frac{e^{2x}-1}{e^{2x}+1}=1-\frac{2}{e^{2x}+1}=1-\frac{2e^{-2x}}{1+e^{-2x}} \).
2. Integrating: \( x+\log(1+e^{-2x})+c'=\log(e^{2x}+1)-x+c \).

Teacher's Note:
1. Divide numerator and denominator by e to the 2x.
2. Use log(1 + e to the minus 2x) = log(e to the 2x + 1) minus 2x.

Answer: (a) \( \log{\left(e^{x} - e^{- x} \right)}+c \)

Explanation:
1. Divide numerator and denominator by \( e^x \): the integrand is \( \frac{e^x+e^{-x}}{e^x-e^{-x}} \).
2. The numerator is the derivative of the denominator, so the integral is \( \log(e^x-e^{-x})+c \).

Teacher's Note:
1. This is the hyperbolic cotangent, whose integral is log of hyperbolic sine.
2. The function in the log must be the denominator.

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

Explanation:
1. Multiply numerator and denominator by \( e^x \): \( \frac{e^x}{e^{2x}+1} \).
2. Put \( t=e^x \): \( \int\frac{dt}{t^2+1}=\tan^{-1}(e^x)+c \).

Teacher's Note:
1. The substitution t = e to the x gives the arctan form.
2. Option (d) integrates a different function.

Answer: (a) \( - \frac{1}{e^{x} + 1}+c \)

Explanation:
1. \( (1+e^{-x})=\frac{e^x+1}{e^x} \), so the integrand is \( \frac{e^x}{(1+e^x)^2} \).
2. Put \( t=1+e^x \): \( \int t^{-2}dt=-\frac{1}{1+e^x}+c \).

Teacher's Note:
1. Simplify the second bracket to get a perfect square in the denominator.
2. The numerator e to the x is the derivative of the base.