Question

Solve:
$I=\int \frac{dx}{\left(x+1\right)\sqrt{1-x^2}}$ .

Answer 1

$I=\int \frac{dx}{(x+1)\sqrt{1-x^2}}=I=\int \frac{dx}{(x+1)\sqrt{(1+x)(1-x)}}$

$I=\int \frac{dx}{{(x+1)}^2}\left(\sqrt{\frac{(1+x)}{(1-x)}}\right)$

put $\sqrt{\frac{1-x}{1+x}}=t\Rightarrow \frac{1-x}{1+x}=t^2$

on differentiation

$\frac{[-1(1+x)-1(1-x)]}{{(1+x)}^2}dx=2tdt$

$\Rightarrow \frac{(-1-x-1+x)}{{(1+x)}^2}dx=2tdt$

$\Rightarrow \frac{dx}{{(1+x)}^2}=-tdt$

$\therefore I=\int tx-tdt=-\int t^{2}dt$

$I=\frac{-t^3}{3}+c=-\frac{1}{3}\left(\sqrt{\frac{(1-x)}{(1+x)}}\right)$

$I=\int (-tdt)\times \frac{1}{t}=\int -dt$

$\Rightarrow I=-t+c=-\sqrt{\frac{1-x}{1+x}}+c$

$\therefore \int \frac{dx}{(x+1)\sqrt{1-x^2}}=-\sqrt{\frac{1-x}{1+x}}+c$