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JEE Main 201411 AprilOnline
Exam
JEE Main
Year
2014
Date
11 April
Mode
Online
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2014 · 11 April · Online · Mathematics · Indefinite Integration

The integral ∫ x ^ -1 ((1-x^2)/(1+x^2)) d x(x>0) is equal to:

The integral ∫xcos⁡−1(1−x21+x2)dx(x>0)\int x \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x(x>0) is equal to:

Options

  1. A−x+(1+x2)tan⁡−1x+c-x+\left(1+x^2\right) \tan ^{-1} x+c
  2. Bx−(1+x2)cot⁡−1x+cx-\left(1+x^2\right) \cot ^{-1} x+c
  3. C−x+(1+x2)cot⁡−1x+c-x+\left(1+x^2\right) \cot ^{-1} x+c
  4. Dx−(1+x2)tan⁡−1x+cx-\left(1+x^2\right) \tan ^{-1} x+c

Correct answer

A. −x+(1+x2)tan⁡−1x+c-x+\left(1+x^2\right) \tan ^{-1} x+c

Step-by-step solution

Let I=∫xcos⁡−1(1−x21+x2)dx\mathrm{I}=\int x \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x \therefore I=2∫II \mathrm{I}=2 \int_{\text{II }} x ⋅tan⁡−1\cdot \tan ^{-1} x d x Applying Integration by parts I=2[tan⁡−1x∫xdx−∫(ddx(tan⁡−1x)∫xdx)dx]I=2[x22tan⁡−1x−∫11+x2×x22dx]+cI=2[x22tan⁡−1x−12∫x2+1−1x2+1dx]+cI=2[x22tan⁡−1x−12∫x2+1x2+1dx+12∫11+x2dx]+c\begin{array}{r}\mathrm{I}=2\left[\tan ^{-1} x \int x d x-\int\left(\frac{d}{d x}\left(\tan ^{-1} x\right)\int x d x\right) d x\right] \\ \mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\int \frac{1}{1+x^2} \times \frac{x^2}{2} d x\right]+c \\ \mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2+1-1}{x^2+1} d x\right]+c \\ \mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2+1}{x^2+1} d x+\frac{1}{2} \int \frac{1}{1+x^2} d x\right]+c \end{array}I=2[x22tan⁡−1x−12∫1.dx+12tan⁡−1x]+cI=2[x22tan⁡−1x−x2+12tan⁡−1x]+cI=x2tan⁡−1x+tan⁡−1x−x+c or I=−x+(x2+1)tan⁡−1x+c\begin{aligned}\mathrm{I}&=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int1 . d x+\frac{1}{2} \tan ^{-1} x\right]+c \\ \mathrm{I}&=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x\right]+c \\ \mathrm{I}&=x² \tan ^{-1} x+\tan ^{-1} x-x+c \\ & \text{ or } I=-x+\left( x^2+1\right) \tan ^{-1} x+c \end{aligned}
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