Answer
$$\pi$$
Work Step by Step
Given $$ \int_{0}^{\infty} \frac{ d x}{(x+1)\sqrt{x}}$$
Let
$$u=\sqrt{x} ,\ \ \to du=\frac{dx}{2\sqrt{x}} $$
Then
\begin{align*}
\int_{0}^{\infty} \frac{ d x}{(x+1)\sqrt{x}}&=\int_{0}^{\infty} \frac{2 d u}{u^{2}+1}\\
&=\int_{0}^{\infty} \frac{2 d u}{u^{2}+1}\\
&=\lim _{b \rightarrow \infty} \int_{0}^{b} \frac{2 \mathrm{du}}{u^{2}+1}\\
&=\lim _{b \rightarrow \infty}\left[2 \tan ^{-1} u\right]_{0}^{b}\\
&=\lim _{b \rightarrow \infty}\left(2 \tan ^{-1} b-2 \tan ^{-1} 0\right)\\
&=\pi
\end{align*}
