Answer
$(x+1) \arctan\sqrt{x}-\sqrt{x}+C$
Work Step by Step
Strategy: see if we can lose the square root...$\left[\begin{array}{ll}
t=\sqrt{x} & t^{2}=x\\
& 2tdt=dx
\end{array}\right]$
$I= \displaystyle \int \arctan \displaystyle \sqrt{x}dx=\int \arctan t(2tdt)= \displaystyle \int u(dv),\qquad $...by parts
$\left[\begin{array}{ll}
u= \arctan t & dv=2tdt\\
du=\dfrac{1}{1+t^{2}}dt & v=t^{2}
\end{array}\right]$
$=uv-\displaystyle \int vdu$
$=t^{2} \displaystyle \arctan t-\int\frac{t^{2}}{1+t^{2}}dt$
$=t^{2} \displaystyle \arctan t-\int\frac{1+t^{2}-1}{1+t^{2}}dt$
$=t^{2} \displaystyle \arctan t-\int(1-\frac{1}{1+t^{2}})dt$
$=t^{2} \displaystyle \arctan t-\int dt+\int\frac{1}{1+t^{2}}dt$
both integrals are in the table: 1. and 17.
$=t^{2}\arctan t-t+ \arctan t+C$
$=(t^{2}+1) \arctan t-t+C$
... bring x back
$=(x+1) \arctan\sqrt{x}-\sqrt{x}+C$