Question

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

Answer 1

\[\int\frac{dx}{\sqrt{\tan^{- 1} x} \left( 1 + x^2 \right)}\]
\[Let \tan^{- 1} x = t\]
\[ \Rightarrow \frac{1}{1 + x^2} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{1 + x^2}dx = dt\]

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

\[ = \int \frac{dt}{\sqrt{t}}\]
\[ = \int t^{- \frac{1}{2}} dt\]

\[ = \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C\]

\[ = 2 \sqrt{t} + C\]
\[ = 2 \sqrt{\tan^{- 1} x} + C\]