<s>[INST] Evaluating $ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $ I would like to evaluate: $$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $$
$$ \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}=\frac{\sqrt{1-x}+\sqrt{1+x}-2}{2(\sqrt{1-x^2}-1)} $$
The substitution $ x \rightarrow \sin(x) $ or $ \cos(x) $ can only simplify the denominator, and $ x \rightarrow \sqrt{1+x}$ or $ \sqrt{1-x} $ is also useless...
Can you help me find a useful substitution?
$$ x=\cos(2t) $$
$$  \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx=-\int {\frac{\sqrt{2}\sin(t)\cos(t)}{\sqrt{2}+\sin(t)+\cos(t)}}\mathrm dt $$
$$ u=\tan(t/2) $$
$$ -4\sqrt{2}\int \frac{u(1-u^2)}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})}\mathrm du $$
But now it looks even more complicated... ?
 [/INST] Just to give a complete answer using the method of the square. Using the easily proved equation
$$\left ( \sqrt{1+x} + \sqrt{1-x} \right )^2 = 2\left ( 1+ \sqrt{1-x^2} \right )$$
one can get 
\begin{align*} 
\int_{-1}^{1} \frac{\mathrm{d}x}{\sqrt{1+x} + \sqrt{1-x} + 2} &= \int_{-1}^{1} \frac{\mathrm{d}x}{\sqrt{2\left (1+\sqrt{1-x^2} \right )}+2} \\ 
&\!\!\!\!\!\overset{x = \sin u}{=\! =\! =\! =\! =\!} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{\sqrt{2\left ( 1+\cos u \right )}+2}\, \mathrm{d}u\\ 
&=\int_{-\pi/2}^{\pi/2} \frac{\cos u}{\sqrt{4 \cos^2 \frac{u}{2}}+2} \, \mathrm{d}u \\ 
&= \int_{-\pi/2}^{\pi/2} \frac{\cos u}{2 \cos \frac{u}{2} + 2} \, \mathrm{d} u \\ 
&= \frac{1}{2} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{1+ \cos \frac{u}{2}} \, \mathrm{d}u \\ &= \frac{1}{4} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{\cos^2 \frac{u}{4}} \, \mathrm{d}u \\ &= {\require{cancel}\cancelto{0}{\frac{1}{4} \left [ 4 \tan \frac{u}{4} \cos u \right ]_{-\pi/2}^{\pi/2}}} + \int_{-\pi/2}^{\pi/2} \tan \frac{u}{4} \sin u \, \mathrm{d}u \end{align*}
The last integral can be dealt as follows:
\begin{align*} 
\int_{-\pi/2}^{\pi/2} \tan \frac{u}{4} \sin u \, \mathrm{d}u \;  &\overset{y=u/4}{=\! =\! =\! =\!} \;  4 \int_{-\pi/8}^{\pi/8} \tan y \sin 4y \, \mathrm{d}y\\ 
&=4 \int_{-\pi/8}^{\pi/8} \tan y \left ( 4 \sin y \cos^3 y - 4 \sin^3 y \cos y \right )\, \mathrm{d}y \\ 
&=4 \int_{-\pi/8}^{\pi/8} \left ( 4 \sin^2 y \cos^2 y - 4 \sin^4 y \right )\, \mathrm{d}y \\ 
&= 16 \int_{-\pi/8}^{\pi/8} \left ( \sin y \cos y \right )^2 \, \mathrm{d} y - 16 \int_{-\pi/8}^{\pi/8} \sin^4 y \, \mathrm{d}y \\ 
&= \left (\frac{\pi}{2} - 1 \right ) + \left (-1 + 4 \sqrt{2} - \frac{3\pi}{2} \right ) \\ 
&= 4\sqrt{2} - \pi - 2 
\end{align*}
Basically as in @RE60K's answer. The results arise in view of the identities:
$$\sin^4 x = \frac{1}{8} \left ( 3 + \cos 4x - 4 \cos 2x \right )$$
and
\begin{align*} 
\int_{-\pi/8}^{\pi/8} \sin^2 x \cos^2 x \, \mathrm{d}x &= \int_{-\pi/8}^{\pi/8} \left ( \sin x \cos x \right )^2 \, \mathrm{d} x\\ 
&=\int_{-\pi/8}^{\pi/8} \left ( \frac{\sin 2x}{2} \right )^2 \, \mathrm{d}x\\ 
&= \frac{1}{4} \int_{-\pi/8}^{\pi/8} \sin^2 2x \, \mathrm{d}x\\ 
&=\frac{1}{8}\int_{-\pi/8}^{\pi/8} \left ( 1- \cos 4x \right ) \, \mathrm{d}x \\ 
&= \frac{1}{8} \left ( \frac{\pi}{4} - \frac{1}{2} \right ) \\
&= \frac{\pi}{32} - \frac{1}{16}
\end{align*}
I know I'm 9 years late but I wanted to present the method of squares.
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