Answer
$$\frac{\pi ^{\frac{3}{2}}}{12}$$
Work Step by Step
Given $$ \int_{0}^{1}\frac{\sqrt{\tan^{-1}x}}{1+x^2}dx$$
Let $u=\tan^{-1}x \ \ \Rightarrow \ \ du =\dfrac{1}{1+x^2}dx $, and at $x=0 \to u=0$, $x=1 \to u=\pi/4$, then
\begin{align*}
\int_{0}^{1}\frac{\sqrt{\tan^{-1}x}}{1+x^2}dx&= \int_{0}^{\pi/4} \sqrt{u}du\\
&=\frac{2}{3}u^{3/2}\bigg|_{0}^{\pi/4}\\
&=\frac{2}{3}\left(\frac{\pi}{4}\right)^{3/2}\\
&=\frac{\pi ^{\frac{3}{2}}}{12}
\end{align*}
