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

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 $y\rightarrow y\pm 1$)
