<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] 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 $\arccos$.
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