Evaluation of $\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$ 
Evaluation of $$\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$

$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$
Put $x=\cos 2 \theta\;,$ Then $dx = -2\sin 2 \theta d\theta$ and Changing Limit, We get
$$I = \int_{0}^{\frac{\pi}{4}}\frac{2\sin 2 \theta}{\sqrt{2}\cos \theta+\sqrt{2}\sin \theta+2}d\theta$$
So $$I = \int_{0}^{\frac{\pi}{4}}\frac{\sin 2 \theta}{\cos\left(\theta-\frac{\pi}{4}\right)+1}d\theta$$
Now Put $\displaystyle \theta-\frac{\pi}{4}=t\;,$ Then $d\theta = dt$ and changing limits
So $$I = \int_{-\frac{\pi}{4}}^{0}\frac{\cos 2t}{\cos t+1}dt=2\int_{-\frac{\pi}{4}}^{0}\frac{2\cos^2 t-2+1}{\cos t+1}dt$$
after that we can solve easily,
My question is can we solve it without Using Trig substution,
Plz explain me
Thanks
 A: I'm not sure this is the most straight forward way, but here it is anyway.
Start with the substitution $\sqrt{1-x}\mapsto x $. The integral transforms into
$$I=\int_0^1 \frac{2 x}{2+x+\sqrt{2-x^2}}dx.$$
Next, rationalize the denominator by multiplying the numerator and the denominator by $2+x-\sqrt{2-x^2}.$
This turns out to work very well, because $\,\,(2+x)^2-(2-x^2)=2(1+x)^2.$
So we get that the integral equals 
$$I=\int_0^1 \frac{x(2+x-\sqrt{2-x^2})}{(1+x)^2}dx.$$
Finally, integrate by parts (note that $\displaystyle \frac1{(1+x)^2}=-\frac{d}{dx} \frac1{1+x}$):
$$I=-\frac{x(2+x-\sqrt{2-x^2})}{1+x}\Bigg{|}_0^1+\int_0^1 2\left(1+\frac{x-1}{\sqrt{2-x^2}}\right)dx
\\\\=-1+2-2\left(\sqrt{2-x^2}+\sin^{-1}\frac{x}{\sqrt{2}}\right)\Bigg{|}_0^1
\\\\=2\sqrt{2}-1-\frac{\pi}{2}.$$
A: multiply the integrand by  
$$
1=\frac{\left(-\sqrt{1-x}-\sqrt{x+1}+2\right) \left(\sqrt{1-x^2}+1\right)}{\left(-\sqrt{1-x}-\sqrt{x+1}+2\right) \left(\sqrt{1-x^2}+1\right)}
$$
doing the alegbra correctly this indeed eliminates the roots in the denominator and we end up with
$$
I=\frac{1}{2}\int_{0}^{1}\frac{-\sqrt{1-x} x+\sqrt{1+x} x-2 \sqrt{1-x}+2 \sqrt{1-x^2}-2 \sqrt{x+1}+2}{x^2}dx=\\
$$
taking into account a proper limiting procedure for $x\rightarrow 0,1$, this integral can be split in elementary pieces (solvable by integration by parts and/or subs of the form $y\rightarrow y\pm 1$)
