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).

By letting $x=\frac{z-1}{z+1}$ the given integral boils down to
$$ 2\int \frac{\arctan\sqrt{z}}{(z+1)^2}\,dz\stackrel{z\mapsto u^2}{=}4\int\frac{u\arctan u}{(u^2+1)^2}\,du\stackrel{\text{IBP}}{=}\frac{u-(1-u^2)\arctan u}{u^2+1}. $$
Your method leads to the same solution by noticing that
$$\frac{d}{dx}\,\arctan\sqrt{\frac{1+x}{1-x}}=\frac{\frac{d}{dx}\sqrt{\frac{1+x}{1-x}}}{1+\frac{1+x}{1-x}}=\frac{1-x}{2}\cdot\frac{\frac{d}{dx}\frac{1+x}{1-x}}{2\sqrt{\frac{1+x}{1-x}}}=\frac{1-x}{2}\cdot\frac{\frac{2}{(1-x)^2}}{2\sqrt{\frac{1+x}{1-x}}}$$
simplifies to $\frac{1}{2\sqrt{1-x^2}}$.
