Answer
$$
\int \frac{d x}{\sqrt{x}(1+x)} =2 \tan ^{-1} \sqrt{x}+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\int \frac{d x}{\sqrt{x}(1+x)}
$$
Let $u=\sqrt{x}.$ Then $d u=\frac{d x}{2 \sqrt{x}} $. and, the Substitution Rule gives
$$
\begin{aligned}
\int \frac{d x}{\sqrt{x}(1+x)} &=\int \frac{2 d u}{1+u^{2}} \\
&=2 \tan ^{-1} u+C\\
&=2 \tan ^{-1} \sqrt{x}+C
\end{aligned}
$$
where $C$ is an arbitrary constant.