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

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 $\displaystyle \frac1{(1+x)^2}=-\frac{d}{dx} \frac1{1+x}$):
$$I=-\frac{x(2+x-\sqrt{2-x^2})}{1+x}\Bigg{|}_0^1+\int_0^1 2\left(1+\frac{x-1}{\sqrt{2-x^2}}\right)dx
\\\\=-1+2-2\left(\sqrt{2-x^2}+\sin^{-1}\frac{x}{\sqrt{2}}\right)\Bigg{|}_0^1
\\\\=2\sqrt{2}-1-\frac{\pi}{2}.$$
