Question

Evaluate: ${\int }_0^{1}x({\tan }^{-1}x{)}^{2}dx$

Answer 1

let $I={\int }_0^{1}x{\left({\tan }^{-1}x\right)}^{2}dx={\left({\tan }^{-1}x\right)}^2{\int }_0^{1}xdx-{\int }_0^1\left\{\frac{d}{dx}{\left({\tan }^{-1}x\right)}^2{\int }_0^{1}xdx\right\}dx$

$={\left[{\left({\tan }^{-1}x\right)}^2\frac{x^2}{2}\right]}_0^1-{\int }_0^1\left\{2{\tan }^{-1}x\frac{d}{dx}\left({\tan }^{-1}x\right){\left(\frac{x^2}{2}\right)}_0^1\right\}dx$

$=[{\left({\tan }^{-1}1\right)}^2\left(\frac{1^2}{2}\right)-{\tan }^{-1}0\left(\frac{0^2}{2}\right)]-{\int }_0^1\left\{2{\tan }^{-1}x\frac{1}{1+x^2}\left(\frac{1}{2}\right)\right\}dx$

$=[{\left(\frac{\pi }{4}\right)}^2\left(\frac{1}{2}\right)-0]-{\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx$

$=\frac{1}{2}\left(\frac{{\pi }^2}{16}\right)-{\int }_0^1\frac{{\tan }^{-1}x}{{1+x}^2}dx\_\_\_\_\_\left(1\right)$

now,

${\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx={\int }_0^{\frac{\pi }{4}}tdt\left[\begin{array}{c}let{\tan }^{-1}x=t\\ \frac{1}{1+x^2}dx=dt\end{array}\right]$

$=\left(\frac{t^2}{2}\right)\begin{array}{c}\frac{x}{4}\\ 0\end{array}=\frac{1}{2}\left(\frac{{\pi }^2}{16}\right)=0$

$\therefore$From $\left(1\right),I=\frac{1}{2}\left(\frac{{\pi }^2}{16}\right)-\frac{1}{2}\left(\frac{{\pi }^2}{16}\right)$

$\therefore {\int }_0^1{\left({\tan }^{-1}x\right)}^{2}dx=0$ answer