JEE Mathematics Indefinite Integration MCQs Set 08

Answer: (a) \( \frac{x^{3}}{9} - x + \left(- \frac{x^{3}}{3} + x\right) \log{\left(x \right)}+c \)

Explanation:
1. By parts with \( u=\log x \) and \( dv=(1-x^2)dx \): \( \int=\left(x-\frac{x^3}{3}\right)\log x-\int\left(x-\frac{x^3}{3}\right)\frac1x dx \).
2. \( \int\left(1-\frac{x^2}{3}\right)dx=x-\frac{x^3}{9} \), so the result is \( \left(x-\frac{x^3}{3}\right)\log x-\left(x-\frac{x^3}{9}\right)+c \).

Teacher's Note:
1. Take log x as the first function in by parts.
2. Simplify the second integral before integrating it.

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

Explanation:
1. By parts with \( u=\log(x^2+a^2) \), \( dv=\frac{dx}{x^2} \): \( -\frac{1}{x}\log(x^2+a^2)+\int\frac1x\cdot\frac{2x}{x^2+a^2}dx \).
2. \( 2\int\frac{dx}{x^2+a^2}=\frac2a\tan^{-1}\frac xa \), giving \( -\frac1x\log(x^2+a^2)+\frac2a\tan^{-1}\frac xa+c \).

Teacher's Note:
1. Integrating 1 over x squared gives minus 1 over x.
2. The leftover rational integral is the standard arctangent form.

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

Explanation:
1. By parts: \( -\frac{\log x}{2x^2}+\frac12\int\frac{dx}{x^3} \).
2. \( \frac12\int x^{-3}dx=-\frac{1}{4x^2} \), so the result is \( -\frac{1}{4x^2}(2\log x+1)+C \).

Teacher's Note:
1. Take log x as u and x to the power minus 3 as dv.
2. Combine the two terms over the common denominator 4 x squared.

Answer: (b) \( \frac{x^{4} \log{\left(x \right)}}{4} - \frac{x^{4}}{16}+c \)

Explanation:
1. By parts: \( \frac{x^4}{4}\log x-\frac14\int x^3dx \).
2. \( =\frac{x^4}{4}\log x-\frac{x^4}{16}=\frac{1}{16}\left[4x^4\log x-x^4\right]+C \).

Teacher's Note:
1. Differentiate log x and integrate the power of x.
2. Writing over 16 gives the standard form.

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

Explanation:
1. By parts: \( \frac{x^2}{2}\sin^{-1}x-\frac12\int\frac{x^2}{\sqrt{1-x^2}}dx \).
2. With \( x=\sin\theta \): \( \int\frac{x^2}{\sqrt{1-x^2}}dx=\frac12\sin^{-1}x-\frac x2\sqrt{1-x^2} \).
3. So the result is \( \left(\frac{x^2}{2}-\frac14\right)\sin^{-1}x+\frac x4\sqrt{1-x^2}+c \).

Teacher's Note:
1. Take the inverse function as u.
2. The remaining integral is done by x = sin theta.

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

Explanation:
1. Put \( x=e^t \): \( \int e^t\cos t\,dt=\frac{e^t}{2}(\cos t+\sin t) \).
2. So the integral is \( \frac x2\{\cos(\log x)+\sin(\log x)\}+c \).

Teacher's Note:
1. The substitution x = e to the t turns the integral into the standard e to the t cos t form.
2. Remember the factor one half.

Answer: (c) \( \dfrac{2x-1}{4} \)

Explanation:
1. By parts: \( \int xe^{2x}dx=\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}=e^{2x}\cdot\frac{2x-1}{4} \).
2. So \( f(x)=\frac{2x-1}{4} \).

Teacher's Note:
1. Integrate e to the 2x as e to the 2x over 2.
2. Take e to the 2x outside as a common factor.

Answer: (c) \( x\sin x+2 \)

Explanation:
1. \( x\cos x+\sin x=\frac{d}{dx}(x\sin x) \), so \( f(x)=x\sin x+c \).
2. \( f(0)=2 \) gives \( c=2 \), so \( f(x)=x\sin x+2 \).

Teacher's Note:
1. Spot the product rule pattern in the derivative.
2. Use the given value to find the constant.

Answer: (d) \( x(\log_{10}x-\log_{10}e)+c \)

Explanation:
1. \( \log_{10}x=\frac{\log x}{\log10} \), so the integral is \( \frac{x\log x-x}{\log10}+c \).
2. Since \( \frac{1}{\log 10}=\log_{10}e \), this equals \( x(\log_{10}x-\log_{10}e)+c \).

Teacher's Note:
1. Convert to natural logarithms first.
2. The constant 1 over log 10 is log of e to base 10.

Answer: (b) \( \frac{\left(2 \sin{\left(3 x \right)} - 3 \cos{\left(3 x \right)}\right) e^{2 x}}{13}+c \)

Explanation:
1. Use \( \int e^{ax}\sin bx\,dx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx) \) with \( a=2,\ b=3 \).
2. The result is \( \frac{e^{2x}}{13}(2\sin3x-3\cos3x)+c \).

Teacher's Note:
1. Apply by parts twice and solve for the integral.
2. a squared plus b squared is 13.

Answer: (c) \( 5 \)

Explanation:
1. \( I=\frac{e^x}{1^2+2^2}(\sin2x-2\cos2x)+c=\frac{e^x}{5}(\sin2x-2\cos2x)+c \).
2. So \( 5I=e^x(\sin2x-2\cos2x)+\text{const.} \) and \( K=5 \).

Teacher's Note:
1. Use the standard formula with a = 1 and b = 2.
2. Multiply the result by the denominator to read off K.

Answer: (c) \( g(x)f(x)+c \)

Explanation:
1. Differentiating \( \int g\,dx=g \) gives \( g'(x)=g(x) \).
2. So \( g(f+f')=g'f+gf'=\frac{d}{dx}\left(g f\right) \), and the integral is \( g(x)f(x)+c \).

Teacher's Note:
1. The condition g dash equals g lets you replace g by g dash.
2. Then the integrand is the derivative of a product.

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

Explanation:
1. By parts: \( x(\log x)^2-2\int\log x\,dx \).
2. \( =x(\log x)^2-2(x\log x-x)=x(\log x)^2-2x\log x+2x+c \).

Teacher's Note:
1. Apply by parts twice.
2. The sign of the last term is plus 2x.

Answer: (d) None of these

Explanation:
1. By parts: \( \int\ln(x^2+x)dx=x\ln(x^2+x)-\int\frac{x(2x+1)}{x^2+x}dx=x\ln(x^2+x)-\int\frac{2x+1}{x+1}dx \).
2. \( \frac{2x+1}{x+1}=2-\frac{1}{x+1} \), so \( A=-2x+\ln(x+1)+\text{const.} \).
3. This matches none of options (a), (b) or (c).

Teacher's Note:
1. Cancel the common factor x in the second integral.
2. Divide 2x + 1 by x + 1 to split it.

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

Explanation:
1. By parts: \( -\frac{\log x}{x+1}+\int\frac{dx}{x(x+1)} \).
2. \( \frac{1}{x(x+1)}=\frac1x-\frac1{x+1} \), so the result is \( -\frac{\log x}{x+1}+\log x-\log(x+1)+c \).

Teacher's Note:
1. Integrate 1 over (x + 1) squared to get minus 1 over (x + 1).
2. Use partial fractions for the remaining integral.

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

Explanation:
1. By parts: \( x\tan^{-1}x-\int\frac{x}{1+x^2}dx \).
2. \( =x\tan^{-1}x-\frac12\log(1+x^2)+c \).

Teacher's Note:
1. Treat the integrand as 1 times tan inverse x.
2. The second integral is of the form f dash over f.

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

Explanation:
1. By parts: \( \frac{x^2}{2}\tan^{-1}x-\frac12\int\frac{x^2}{1+x^2}dx \).
2. \( \frac{x^2}{1+x^2}=1-\frac{1}{1+x^2} \), so the result is \( \frac{x^2}{2}\tan^{-1}x-\frac x2+\frac12\tan^{-1}x+c=\frac12(x^2+1)\tan^{-1}x-\frac x2+c \).

Teacher's Note:
1. Divide x squared by 1 + x squared before integrating.
2. Collect the two arctangent terms.

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

Explanation:
1. Take \( f=\sin^{-1}\frac xa \); then \( f'=\frac{1}{\sqrt{a^2-x^2}} \).
2. So the integral is \( e^x\sin^{-1}\frac xa+c \).

Teacher's Note:
1. Check that the second term is the derivative of the first.
2. Then apply the e to the x rule.

Answer: (a) \( e^{x} \sin^{2}{\left(x \right)}+c \)

Explanation:
1. The integrand is \( e^x(\sin^2x+2\sin x\cos x)=e^x(\sin^2x+\sin2x) \).
2. With \( f=\sin^2x \), \( f'=\sin2x \), so the integral is \( e^x\sin^2x+c \).

Teacher's Note:
1. Expand the product and use sin 2x = 2 sin x cos x.
2. Identify f and f dash.

Answer: (a) \( e^xf(x)+c \)

Explanation:
1. \( \frac{d}{dx}\left(e^xf(x)\right)=e^x f(x)+e^xf'(x) \).
2. Hence the integral is \( e^xf(x)+c \).

Teacher's Note:
1. This is the basic identity behind the e to the x method.
2. Differentiating the result confirms it.

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

Explanation:
1. \( \frac{x-1}{(x+1)^3}=\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3} \), which is \( f+f' \) for \( f=\frac{1}{(x+1)^2} \).
2. The integral is \( \frac{e^x}{(x+1)^2}+c \).

Teacher's Note:
1. Write x - 1 as (x + 1) - 2 in the numerator.
2. Then use the f plus f dash rule.

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

Explanation:
1. \( \frac{1+\sin x}{1+\cos x}=\frac{1}{2\cos^2\frac x2}+\frac{2\sin\frac x2\cos\frac x2}{2\cos^2\frac x2}=\frac12\sec^2\frac x2+\tan\frac x2 \).
2. With \( f=\tan\frac x2 \), \( f'=\frac12\sec^2\frac x2 \), so the integral is \( e^x\tan\frac x2+C \).

Teacher's Note:
1. Use half angle identities to rewrite the fraction.
2. Identify f and f dash for the e to the x rule.

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

Explanation:
1. \( 1+\tan^2x=\sec^2x \), so the integrand is \( e^x(\tan x+\sec^2x) \).
2. With \( f=\tan x \), \( f'=\sec^2x \), so the integral is \( e^x\tan x+C \).

Teacher's Note:
1. Combine 1 and tan squared into sec squared.
2. Then use the f plus f dash rule.

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

Explanation:
1. Try \( \frac{d}{dx}\left(xe^{x+1/x}\right)=e^{x+1/x}\left[1+x\left(1-\frac{1}{x^2}\right)\right] \).
2. This equals \( e^{x+1/x}\left(1+x-\frac1x\right) \), so the integral is \( xe^{x+1/x}+C \).

Teacher's Note:
1. Recognise the product of x and an exponential whose exponent is x + 1/x.
2. Verify by differentiating.

Answer: (d) None of these

Explanation:
1. \( \frac{d}{dx}\left(e^{-x}\sec x\right)=e^{-x}(-\sec x+\sec x\tan x)=-e^{-x}(1-\tan x)\sec x \).
2. So the integral is \( -e^{-x}\sec x+C \), which is not among options (a), (b), (c).

Teacher's Note:
1. The factor e to the minus x changes the sign of the result.
2. Always differentiate a candidate to test it.