Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{\left(1+x^2\right)\sqrt{1-x^2}} \\ \mathrm{Putting}x=\frac{1}{t} \\ \Rightarrow dx=-\frac{1}{t^2}dt \\ \therefore I=\int \frac{-{\displaystyle \frac{1}{t^2}}dt}{\left(1+{\displaystyle \frac{1}{t^2}}\right)\sqrt{1-{\displaystyle \frac{1}{t^2}}}} \\ =\int \frac{-{\displaystyle \frac{1}{t^2}}dt}{{\displaystyle \frac{\left(t^2+1\right)}{t^2}\frac{\sqrt{t^2-1}}{t}}} \\ =-\int \frac{tdt}{\left(t^2+1\right)\sqrt{t^2-1}} \\ \mathrm{Again}\mathrm{Putting}{t}^2-1=u^2 \\ \Rightarrow 2tdt=2udu \\ \Rightarrow tdt=udu \\ \therefore I=-\int \frac{udu}{\left(u^2+2\right)u} \\ =-\int \frac{du}{u^2+{\left(\sqrt{2}\right)}^2} \\ =-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)+C \\ =-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{t^2-1}}{\sqrt{2}}\right)+C \\ =-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{\frac{{\displaystyle \frac{1}{x^2}-1}}{2}}\right)+C \\ =-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{\frac{1-x^2}{2x^2}}\right)+C \end{gathered}$$