Answer
$$
\int \frac{\arctan \sqrt{x}}{\sqrt{x}} d x =2 \sqrt{x} \arctan \sqrt{x}-\ln (1+x)+C
$$
Work Step by Step
$$
\int \frac{\arctan \sqrt{x}}{\sqrt{x}} d x
$$
If we make the substitution
$$
u=\sqrt{x}, \quad d u=1 /(2 \sqrt{x}) d x
$$
and we look at the Table of Integrals , we see that the closest entry is number $89$ :
$$
\begin{aligned}
\int \frac{\arctan \sqrt{x}}{\sqrt{x}} d x &=\int \arctan u(2 d u) \quad\left[\begin{array}{c}
u=\sqrt{x}, \\
d u=1 /(2 \sqrt{x}) d x
\end{array}\right] \\
& \stackrel{89}{=} 2\left[u \arctan u-\frac{1}{2} \ln \left(1+u^{2}\right)\right]+C\\
&=2 \sqrt{x} \arctan \sqrt{x}-\ln (1+x)+C
\end{aligned}
$$
