Answer
$\displaystyle \frac{\pi^{3/2}}{12}$
Work Step by Step
$\displaystyle \int\frac{\sqrt{\arctan x}}{1+x^{2}}=\left[\begin{array}{l}
u=\arctan x\\
du=\frac{dx}{1+x^{2}}
\end{array}\right]$
$=\displaystyle \int u^{1/2}du$
$=\displaystyle \frac{2}{3}u^{3/2}+C$
$=\displaystyle \frac{2}{3}(\arctan x)^{3/2}+C$
$\displaystyle \int_{0}^{1}\frac{\sqrt{\arctan x}}{1+x^{2}}dx=\frac{2}{3}\left[(\arctan x)^{3/2}\right]_{0}^{1}$
$=\displaystyle \frac{2}{3}(\frac{\pi}{4})^{3/2}-0$
$=\displaystyle \frac{2}{3}(\frac{\pi^{3/2}}{(2^{2})^{3/2}})$
$=\displaystyle \frac{\pi^{3/2}}{12}$
