Question

\[\int\frac{x^2 \tan^{- 1} x}{1 + x^2} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int \left( \frac{x^2 \tan^{- 1} x}{1 + x^2} \right)dx\]
\[ = \int \left( \frac{x^2 + 1 - 1}{x^2 + 1} \right) \tan^{- 1} \text{ x dx }\]
\[ = \int \left( 1 - \frac{1}{x^2 + 1} \right) \tan^{- 1}\text{  x dx }\]
\[ = \int 1_{II} . \tan^{- 1}_I \text{ x dx } - \int \frac{\tan^{- 1} x}{x^2 + 1} \text{ dx}\]

\[ = \left[ \tan^{- 1} x\int1\text{ dx }- \int\left\{ \frac{d}{dx}\left( \tan^{- 1} x \right)\int1 \text{ dx } \right\} \text{ dx }\right] - \int \frac{\tan^{- 1} x}{x^2 + 1}dx\]
\[ = t\left[ {an}^{- 1} x \times x - \int\frac{x}{1 + x^2}dx \right] - \int\frac{\tan^{- 1} x}{x^2 + 1}dx\]
`  " Putting x"^2" + 1 = t in the first integral and tan"^{- 1}" x = p in the second integral " `
\[ \Rightarrow \text{ 2x dx }= dt \text{ and }\frac{1}{1 + x^2}dx = dp\]
\[ \Rightarrow \text{ x dx } = \frac{dt}{2} \text{ and }\frac{1}{1 + x^2}dx = dp\]
\[ \therefore I = \tan^{- 1} x . x - \frac{1}{2}\int \frac{dt}{t} - \int p . dp\]
\[ = x \tan^{- 1} x - \frac{1}{2}\text{ ln} \left| t \right| - \frac{p^2}{2} + C\]
\[ = x \tan^{- 1} x - \frac{1}{2}\text{ ln }\left| 1 + x^2 \right| - \frac{\left( \tan^{- 1} x \right)^2}{2} + C \left[ \because t = x^2 + 1 \text{ and } p = \tan^{- 1} x \right]\]