Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1\frac{\sqrt{{\tan }^{-1}x}}{1+x^2}\mathit{}\mathit{d}x\mathit{.} \\ \mathrm{Let}{\tan }^{-1}x=t.\mathrm{Then},\frac{1}{1+x^2}dx=dt \\ \mathrm{When}x=0,t=0\mathrm{and}x=1,t=\frac{\mathrm{\pi }}{4} \\ \therefore I={\int }_0^1\frac{\sqrt{{\tan }^{-1}x}}{1+x^2}\mathit{d}x \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{4}}\sqrt{t}dt \\ \Rightarrow I={\left[\frac{2t^{{\displaystyle \frac{3}{2}}}}{3}\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ \Rightarrow I=\frac{2}{3}{\left(\frac{\mathrm{\pi }}{4}\right)}^{\frac{3}{2}} \\ \Rightarrow I=\frac{1}{12}{\mathrm{\pi }}^{\frac{3}{2}} \end{gathered}$$