Integrate $\arctan{\sqrt{\frac{1+x}{1-x}}}$ I use partial integration by letting $f(x)=1$ and $g(x)=\arctan{\sqrt{\frac{1+x}{1-x}}}.$ Using the formula: 
$$\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx,$$
I get
$$\int1\cdot\arctan{\sqrt{\frac{1+x}{1-x}}}dx=x\arctan{\sqrt{\frac{1+x}{1-x}}}-\int\underbrace{x\left(\arctan{\sqrt{\frac{1+x}{1-x}}}\right)'}_{=D}dx.$$
So, the integrand $D$ remains to simplify:
$$D=x\cdot\frac{1}{1+\frac{1+x}{1-x}}\cdot\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot\frac{1}{(x-1)^2} \quad \quad (1).$$
Setting $a=\frac{1+x}{1-x}$ for notation's sake I get
$$D=x\cdot\frac{1}{1+a}\cdot\frac{1}{2\sqrt{a}}\cdot\frac{1}{(x-1)^2}=\frac{x}{(2\sqrt{a}+2a\sqrt{a})(x^2-2x+1)},$$
and I get nowhere. Any tips on how to move on from $(1)?$
NOTE: I don't want other suggestions to solutions, I need help to sort out the arithmetic to the above from equation (1).

Use change of variable
$$\theta=\arctan\sqrt{\frac{1+x}{1-x}}\in[0,\frac{\pi}{2}).$$
Then we have
$$x=\frac{\tan^2\theta-1}{\tan^2\theta+1}=\sin^2\theta-\cos^2\theta=-\cos2\theta.$$
Therefore
\begin{align}
\int\arctan\sqrt{\frac{1+x}{1-x}}dx&=-\int\theta\,d\cos 2\theta=-\theta\cos 2\theta+\int\cos 2\theta d\theta\\
&=-\theta\cos 2\theta+\frac{1}{2}\sin 2\theta+C\\
&=x\arctan\sqrt{\frac{1+x}{1-x}}+\frac{1}{2}\sqrt{1-x^2}+C.
\end{align}
