Question

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

Answer 1

\[\text{ Let I }= \int \tan^{- 1} \sqrt{x} \text{ dx }\]
\[ = \int \frac{\sqrt{x} . \tan^{- 1} \sqrt{x}\text{  dx}}{\sqrt{x}}\]
\[\text{ Let } \sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}} \text{ dx }= dt\]
\[ = \frac{dx}{\sqrt{x}} = 2dt\]
\[ \therefore I = 2\int t_{II} . \tan^{- 1}_I \left( t \right)  \text{  dt }\]
\[ = 2 \left[ \tan^{- 1} t\int\text{  t dt  }- \int\left\{ \frac{d}{dt}\left( \tan^{- 1} t \right)\int \text{ t dt } \right\}dt \right]\]
\[ = 2 \left[ \tan^{- 1} \left( t \right) . \frac{t^2}{2} - \int \frac{1}{1 + t^2} . \frac{t^2}{2}dt \right]\]
\[ = \tan^{- 1} \left( t \right) . t^2 - \int \frac{t^2}{1 + t^2} dt\]
\[ = \tan^{- 1} \left( t \right) . t^2 - \int \left( \frac{1 + t^2 - 1}{1 + t^2} \right)dt\]
\[ = \tan^{- 1} \left( t \right) . t^2 - \int dt + \int\frac{dt}{1 + t^2}\]
\[ = \tan^{- 1} \left( t \right) . t^2 - t + \tan^{- 1} \left( t \right) + C \left( \because \sqrt{x} = t \right)\]
\[ = \tan^{- 1} \left( \sqrt{x} \right) . x - \sqrt{x} + \tan^{- 1} \sqrt{x} + C\]
\[ = \left( x + 1 \right) \tan^{- 1} \sqrt{x} - \sqrt{x} + C\]