JEE Main 2014 · 11 Apr · Shift 1 · Mathematics Q75

MathematicsSingle correct+4 / -1Question 75 of 90

The integral $\displaystyle\int x\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx$ $(x>0)$ is equal to :

-x+\left(1+x^2\right)\tan^{-1}x+c

x-\left(1+x^2\right)\cot^{-1}x+c

-x+\left(1+x^2\right)\cot^{-1}x+c

x-\left(1+x^2\right)\tan^{-1}x+c

Answer and worked solution

Official answer: A

For $x>0$ , $\cos^{-1}\dfrac{1-x^2}{1+x^2}=2\tan^{-1}x$ . Then
$\int2x\tan^{-1}x\,dx=x^2\tan^{-1}x-\int\frac{x^2}{1+x^2}dx=(1+x^2)\tan^{-1}x-x+c.$

JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 75