I=x44tan−1⁡x−14∫x4−1+1x2+1dx=x44tan−1⁡x−14∫x2−1+1x2+1dx etc.

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$\int x^{3} \tan^{- 1} x d x =$

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$[(x^{4} - 1) \tan^{- 1} x - \frac{x^{3}}{3} + x]$

$\frac{1}{2} [(x^{4} - 1) \tan^{- 1} x - \frac{x^{3}}{3} + x]$

$\frac{1}{3} [(x^{4} - 1) \tan^{- 1} x - \frac{x^{3}}{3} + x]$

$\frac{1}{4} [(x^{4} - 1) \tan^{- 1} x - \frac{x^{3}}{3} + x]$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \frac{x^{4}}{4} \tan^{- 1} x - \frac{1}{4} \int \frac{x^{4} - 1 + 1}{x^{2} + 1} d x \\ = \frac{x^{4}}{4} \tan^{- 1} x - \frac{1}{4} \int [(x^{2} - 1) + \frac{1}{x^{2} + 1}] d x etc. \end{array}$

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