Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ $$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$
I've tried substituting $x=2\tan^2y$, and I've got:
$$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$
But I'm not entirely sure this is a good thing as I've been unable to proceed any further from there.

\begin{align}
u & = \arctan \sqrt{\frac x 2} \\[10pt]
du & = \frac{dx/2}{\left(1+ \dfrac x 2\right)2\sqrt{\dfrac x 2}} = \frac{dx}{(2+x)\sqrt{2x}} \\[10pt]
dv & = \frac{dx}{\sqrt{x+2}} \\[10pt]
v & = 2\sqrt{x+2}
\end{align}
\begin{align}
\int u\,dv & = uv - \int v\,du = 2\sqrt{x+2} \arctan \sqrt{\frac x 2} - \int \frac{\sqrt 2\,dx}{\sqrt{x+2}\sqrt x}
\end{align}
Then
$$
\sqrt{x^2+2x} = \sqrt{(x^2+2x + 1) - 1} = \sqrt{(x+1)^2 -1} = \sqrt{\sec^2\theta - 1},\quad\text{etc.}
$$
