Question

\[\int x^3 \tan^{- 1}\text{  x dx }\]

Answer 1

\[\int {x^3}_{II} . \tan^{- 1}_I \text{ x dx }\]
\[ = \tan^{- 1} x \int x^3 dx - \int\left\{ \frac{d}{dx}\left( \tan^{- 1} x \right)\int x^3 dx \right\}dx\]
\[ = \tan^{- 1} x . \frac{x^4}{4} - \int\frac{1}{1 + x^2} \times \frac{x^4}{4}dx\]
\[ = \tan^{- 1} x . \frac{x^4}{4} - \frac{1}{4}\int \frac{x^4 dx}{x^2 + 1}\]
\[ = \tan^{- 1} x . \frac{x^4}{4} - \frac{1}{4}\int \left( \frac{x^4 - 1 + 1}{x^2 + 1} \right)dx\]
\[ = \tan^{- 1} x . \frac{x^4}{4} - \frac{1}{4}\int \left( \frac{x^4 - 1}{x^2 + 1} \right)dx - \frac{1}{4}\int \frac{1}{x^2 + 1}dx\]
\[ = \tan^{- 1} x . \frac{x^4}{4} - \frac{1}{4}\int\frac{\left( x^2 - 1 \right) \left( x^2 + 1 \right)}{\left( x^2 + 1 \right)}dx - \frac{1}{4}\int \frac{1}{x^2 + 1}dx\]
\[ = \tan^{- 1} x . \frac{x^4}{4} - \frac{1}{4}\int \left( x^2 - 1 \right)dx - \frac{1}{4} \tan^{- 1} x + C\]
\[ = \tan^{- 1} x . \frac{x^4}{4} - \frac{1}{4}\left( \frac{x^3}{3} - x \right) - \frac{1}{4} \tan^{- 1} x + C\]
\[ = \left( \frac{x^4 - 1}{4} \right) \tan^{- 1} x - \frac{1}{12}\left( x^3 - 3x \right) + C\]