Question

Find the value of the following integrals:
${\int }_{-1}^{1}x{\tan }^{-1}xdx$

Answer 1

${\int }_{-1}^{1}x{\tan }^{-1}xdx$
$=2{\int }_0^{1}x{\tan }^{-1}xdx$
($\because x{\tan }^{-1}x$ is even function)
$={\left[2\frac{x^2}{2}{\tan }^{-1}x\right]}_0^1-2{\int }_0^1\frac{1}{2}\frac{x^2}{1+x^2}dx$
$=[x^2{\tan }^{-1}x{]}_0^1-{\int }_0^1\frac{x^2+1-1}{1+x^2}dx$
$=[x^2{\tan }^{-1}x{]}_0^1-[x{]}_0^1+[{\tan }^{-1}{]}_0^1$
$=\frac{\pi }{4}-1+\frac{\pi }{4}=\frac{\pi }{2}-1=\frac{\pi -2}{2}$