Question

$\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}x}{1+x^2}dx$

Answer 1

let $ta{n}^{-1}x=t$

$\frac{dta{n}^{-1}x}{dx}=\frac{dt}{dx}$

$\frac{1}{1+x^2}=\frac{dt}{dx}$

$\frac{1}{1+x^2}dx=dt$

Hence limits will be 0 and $\frac{\pi }{4}$

Now Putting the value of $tanddt$

We get,

${\int }_0^{\frac{\pi }{4}}tdt=\frac{t^2}{2}$

$=(\frac{\pi }{4}{)}^2-0$

$=\frac{{\pi }^2}{16}$