Question

${\int }_0^{1}x{\tan }^{-1}xdx$ is equal to :

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{4}+\frac{1}{2}$
• C. $-\frac{\pi }{4}+\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $-\frac{\pi }{4}+\frac{1}{2}$

We have $I={\int }_0^{-1}x{\tan }^{-1}xdx$
Using by parts integration
$I={\left[\frac{x^2}{2}{\tan }^{-1}x\right]}_0^1-{\int }_0^{-1}\frac{x^2}{1+x^2}\left(\frac{1}{2}\right)dx$
$I=\frac{1}{2}.\frac{\pi }{4}-\frac{1}{2}{\int }_0^{-1}(\frac{(1+n^2)dx}{1+n^2}-\frac{dx}{1+n^2})$
$I=-\frac{\pi }{8}-\frac{1}{2}{[x-{\tan }^{-1}x]}_0^{-1}$
$I=-\frac{\pi }{8}-\frac{1}{2}[-1+\frac{\pi }{8}]$
$I=-\frac{\pi }{8}-\frac{\pi }{8}+\frac{1}{2}$
$I=-\frac{\pi }{4}+\frac{1}{2}$