Answer
$\pi/2$
(converges)
Work Step by Step
$\displaystyle \int\frac{dx}{\sqrt{x}+x\sqrt{x}}=\int\frac{dx}{\sqrt{x}(1+x)}=\left[\begin{array}{l}
u=\sqrt{x}\\
du=\frac{1}{2\sqrt{x}}dx
\end{array}\right]$
$=2\displaystyle \int\frac{du}{1+u^{2}}$
$=2\arctan u+C$
$=2\arctan\sqrt{x}+C$
The integral is a type I improper integral, so
$I=\displaystyle \int_{1}^{\infty}\frac{dx}{\sqrt{x}+x\sqrt{x}}=\lim_{t\rightarrow\infty}\left[2\arctan\sqrt{x}\right]_{1}^{t}$
$=2\displaystyle \lim_{t\rightarrow\infty}\arctan\sqrt{t}-2\lim_{t\rightarrow\infty}\arctan\sqrt{1}$
$=\pi-\pi/2$
$=\pi/2$
(converges)
