∫x3−1x3+xdx=∫x3+x−x−1x3+xdx=∫1−x+1xx2+1dx=∫1−1x+x−1x2+1dx=x−log⁡x+12log⁡x2+1−Tan−1⁡x+c

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

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$x - \log x + \log (x^{2} + 1) - {Tan}^{- 1} x + c$

$x - \log x + \frac{1}{2} \log (x^{2} + 1) - {Tan}^{- 1} x + c$

$x + \log x + \log (x^{2} + 1) - {Tan}^{- 1} x + c$

$x + \log x + \frac{1}{2} \log (x^{2} + 1) - {Tan}^{- 1} x + c$

answer is B.

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Detailed Solution

$\begin{array}{l} \int [\frac{x^{3} - 1}{x^{3} + x}] d x = \int \frac{x^{3} + x - x - 1}{x^{3} + x} d x = \int [1 - \frac{x + 1}{x (x^{2} + 1)}] d x = \int [1 - \frac{1}{x} + \frac{x - 1}{x^{2} + 1}] d x \\ = x - \log x + \frac{1}{2} \log (x^{2} + 1) - {Tan}^{- 1} x + c \end{array}$