Evaluate $\int _0^1\:\frac{2-x^2}{(1+x)\sqrt{1-x^2}} \, dx$ 
Evaluate $$\int _0^1\:\frac{2-x^2}{(1+x)\sqrt{1-x^2}}dx$$

$$
\int _0^1\:\frac{2-x^2}{(1+x)\sqrt{1-x^2}} \, dx = \int_0^1 \bigg[\sqrt{\frac{1+x}{1-x}}+\frac{1}{(1-x)^{3/2}\sqrt{1+x}}\bigg] \, dx=\int_0^1[f(x)+f'(x)]\,dx
$$
Does making it into the above final form helps somehow solve the given definite integral ?
Note: The solution given is $\pi/2$

$$x=:\sin u\implies dx=\cos u\,du\implies\int_0^1\frac{2-x^2}{(1+x)\sqrt{1-x^2}}dx=\int_0^{\pi/2}\frac{2-\sin^2u}{(1+\sin u)}\,du=$$
$$=\int_0^{\pi/2}\left(\frac1{1+\sin u}+1-\sin u\right)du=\int_0^{\pi/2}\frac{1-\sin u}{\cos^2u}+\left.\frac\pi2+\cos u\right|_0^{\pi/2}=$$
$$=\left.\left(\tan u-\frac1{\cos u}\right)\right|_0^{\pi/2}+\frac\pi2-1=\frac\pi2$$
since
$$\lim_{u\to\frac\pi2^-}\left(\tan u-\frac1{\cos u}\right)=\lim_{u\to\frac\pi2^-}\frac{\sin u-1}{\cos u}\stackrel{\text{L'Hospital}}=0$$
