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).

Inverse trig identities are less familiar than regular trig identities. So perhaps if you invert this, so that regular trig identities would apply, you find something noteworthy:
$$\begin{align}
y&=\arctan\sqrt{\frac{1+x}{1-x}}\\
\tan(y)&=\sqrt{\frac{1+x}{1-x}}\\
\tan^2(y)&=\frac{1+x}{1-x}\\
\tan^2(y)-x\tan^2(y)&=1+x\\
-x-x\tan^2(y)&=1-\tan^2(y)\\
x&=\frac{\tan^2(y)-1}{\tan^2(y)-1}\\
&=\frac{\tan^2(y)-1}{\sec^2(y)}\\
&=\sin^2(y)-\cos^2(y)\\
&x=-\cos(2y)
\end{align}$$
So now you can invert that back, and actually $y=\frac{1}{2}\arccos(-x)$, much easier to work with if you know the antiderivative for $\arccos$.
