Question

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

Answer 1

\[\text{ We  have,} \]
\[I = \int \frac{dx}{\left( 1 + x^2 \right) \sqrt{1 - x^2}}\]
\[\text{ Putting  x }= \frac{1}{t}\]
\[ \Rightarrow dx = - \frac{1}{t^2}dt\]
\[ \therefore I = \int \frac{- \frac{1}{t^2}dt}{\left( 1 + \frac{1}{t^2} \right) \sqrt{1 - \frac{1}{t^2}}}\]
\[ = \int \frac{- \frac{1}{t^2}dt}{\frac{\left( t^2 + 1 \right)}{t^2} \frac{\sqrt{t^2 - 1}}{t}}\]
\[ = - \int \frac{t dt}{\left( t^2 + 1 \right) \sqrt{t^2 - 1}}\]
\[\text{ Again  Putting t}^2 - 1 = u^2 \]
\[ \Rightarrow 2t \text{ dt} = 2u \text{ du}\]
\[ \Rightarrow t \text{ dt} = u \text{ du }\]
\[ \therefore I = - \int \frac{u \text{ du}}{\left( u^2 + 2 \right)u}\]
\[ = - \int \frac{du}{u^2 + \left( \sqrt{2} \right)^2}\]
\[ = - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{u}{\sqrt{2}} \right) + C\]
\[ = - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{\sqrt{t^2 - 1}}{\sqrt{2}} \right) + C\]
\[ = - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \sqrt{\frac{\frac{1}{x^2} - 1}{2}} \right) + C\]
\[ = - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \sqrt{\frac{1 - x^2}{2 x^2}} \right) + C\]