2. \(\displaystyle \int\frac{2\,-\,x^2}{(1\,+\,x)\sqrt{1\,-\,x^2}}\,dx\)

\(\displaystyle \sin(\theta)\,=\,x,\,\cos(\theta)\,d\theta\,=\,dx\)

\(\displaystyle \int\frac{\[2\,-\,\sin^2(\theta)\]\cos(\theta)}{\[1\,+\,\sin(\theta)\]\cos(\theta)}\,d\theta\,=\,\int\frac{1}{1\,+\sin(\theta)}\,+\,1\,-\,\sin(\theta)\,d\theta\,=\,\int\frac{1}{1\,+\,\sin(\theta)}\,d\theta\,+\,\theta\,+\,\cos(\theta)\,+\,C\)

\(\displaystyle \int\frac{1}{1\,+\,\sin(\theta)}\,d\theta\,\Leftrightarrow\,t\,=\,\tan\(\frac{\theta}{2}\)\,\Leftrightarrow\,\int\frac{2}{(1\,+\,t)^2}\,dt\,=\,-2(1\,+\,t)^{-1}\,+\,C\,=\,-2\frac{1}{1\,+\,\tan\(\frac{\theta}{2}\)}\,+\,C\)

\(\displaystyle -2\frac{1}{1\,+\,\tan\(\frac{\theta}{2}\)}\,+\,\theta\,+\,\cos(\theta)\,+\,C\,=\,-2\,\cdot\,\frac{1\,+\,\sqrt{1\,-\,x^2}}{1\,+\,x\,+\,\sqrt{1\,-\,x^2}}\,+\,\sin^{\small{-1}}(x)\,+\,\sqrt{1\,-\,x^2}\,+\,C\,=\,\int\frac{2\,-\,x^2}{(1\,+\,x)\sqrt{1\,-\,x^2}}\,dx\)

I used the

Weierstrass Substitution.