Answer
$$
\int_{0}^{\infty} \frac{d x}{\sqrt{x}(1+x)}
$$
The given improper integral is convergent and its value is
$$
\int_{0}^{\infty} \frac{d x}{\sqrt{x}(1+x)}=\pi
$$
Work Step by Step
$$
\int_{0}^{\infty} \frac{d x}{\sqrt{x}(1+x)}
$$
The given improper integral can be expressed as a sum of improper integrals of Type 2 and Type 1 as follows:
$$
\int_{0}^{\infty} \frac{d x}{\sqrt{x}(1+x)}=\int_{0}^{1} \frac{d x}{\sqrt{x}(1+x)}+\int_{1}^{\infty} \frac{d x}{\sqrt{x}(1+x)}
$$
Observe that the first integral is improper because $ f(x)= \frac{1}{\sqrt{x}(1+x)} $ has the vertical asymptote $ x =0 $. Since the infinite discontinuity occurs at the left endpoint of $ [0,1] $
$$
\begin{split}
\int_{0}^{\infty} \frac{d x}{\sqrt{x}(1+x)} &=\int_{0}^{1} \frac{d x}{\sqrt{x}(1+x)}+\int_{1}^{\infty} \frac{d x}{\sqrt{x}(1+x)} \\
&=\lim _{t \rightarrow 0^{+}} \int_{t}^{1} \frac{d x}{\sqrt{x}(1+x)}+\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{d x}{\sqrt{x}(1+x)} \quad (1)
\end{split}
$$
Now
$$
\begin{split}
\int \frac{d x}{\sqrt{x}(1+x)} &=\int \frac{d x}{\sqrt{x}(1+x)} \\
& \quad\quad\left[\begin{array}{c}{u=\sqrt{x}, x=u^{2}} \\ {d x=2 u d u}\end{array}\right] \\
& =\int \frac{2 u d u}{u\left(1+u^{2}\right)} \\
& =2 \int \frac{d u}{1+u^{2}} \\
&=2 \tan ^{-1} u+C \\
&=2 \tan ^{-1} \sqrt{x}+C
\end{split}
$$
substituting in Eq. (1 ) we obtain that, $$
\begin{split}
\int_{0}^{\infty} \frac{d x}{\sqrt{x}(1+x)} &=\lim _{t \rightarrow 0^{+}} \left[ 2 \tan ^{-1} \sqrt{x} \right]_{t}^{1} +\lim _{t \rightarrow \infty}\left[ 2 \tan ^{-1} \sqrt{x} \right]_{1}^{t} \\
&=\lim _{t \rightarrow 0+}\left[2\left(\frac{\pi}{4}\right)-2 \tan ^{-1} \sqrt{t}\right] +\\
&\quad+\lim _{t \rightarrow \infty}\left[2 \tan ^{-1} \sqrt{t}-2\left(\frac{\pi}{4}\right)\right] \\
&=\frac{\pi}{2}-0+2\left(\frac{\pi}{2}\right)-\frac{\pi}{2}\\
&=\pi
\end{split}
$$
The given improper integral is convergent and its value is
$$
\int_{0}^{\infty} \frac{d x}{\sqrt{x}(1+x)}=\pi
$$
