<s>[INST] 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).
 [/INST] Hint: make the substitution $x = \cos \theta$ with $0 \le \theta \le \pi$.  Then $1 + x = 2 \cos^2(\theta/2)$, $1 - x = 2 \sin^2(\theta/2)$, so $\sqrt{\frac{1+x}{1-x}} = |\cot(\theta/2)|$.  But by the choice of range of $\theta$, $\cot(\theta/2)$ is positive.  So $\arctan \sqrt{\frac{1+x}{1-x}} = \frac{\pi}{2} - \frac{\theta}{2}$.  (Note this if you restrict to working with real numbers only this will work only for $x \in [-1,1)$; however, for $x > 1$ or $x < -1$, the square root is pure imaginary anyway.  If you want to integrate it anyway, you might need other arguments, for example $x = \cosh \theta$ for $x > 1$.)
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