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 $\pi/2$

Let $x=\sin t$. Then,
$$\int _0^1 \frac{2-x^2}{\left(1+x\right)\sqrt{1-x^2}}dx
=\int _0^1\left(\frac{1}{\left(1+x\right)\sqrt{1-x^2}}
+\frac{\sqrt{1-x^2}}{1+x}\right)dx$$
$$=\int_0^{\pi/2} \left( \frac{dt}{1+\sin t}+1-\sin t\right)dt$$
$$=\left(-\frac{\cos t}{1+\sin t} + t + \cos t\right)_0^{\pi/2}=\frac\pi2$$
