Answer
$2\sqrt{x}\tan^{-1}\sqrt{x}-\ln(1+x)+C$
Work Step by Step
The formula to use is
89. $\displaystyle \int\tan^{-1}udu=u\tan^{-1}u-\frac{1}{2}\ln(1+u^{2})+C$
---
$\displaystyle \int\frac{\arctan\sqrt{x}}{\sqrt{x}}dx=\qquad \left[\begin{array}{ll}
u=\sqrt{x} & du=\frac{1}{2\sqrt{x}}dx\\
& \frac{dx}{\sqrt{x}}=2du
\end{array}\right]$
$=2\displaystyle \int\arctan udu$
Apply the formula
$=2[u\displaystyle \tan^{-1}u-\frac{1}{2}\ln(1+u^{2})]+C$
... bring back the variable x
$=2[\displaystyle \sqrt{x}\tan^{-1}\sqrt{x}-\frac{1}{2}\ln(1+x)[+C$
$=2\sqrt{x}\tan^{-1}\sqrt{x}-\ln(1+x)+C$
