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$\int \frac{x^{3}}{x + 1} d x is equal to$

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$x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \ln | 1 + x | + C$

$x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \ln | 1 + x | + C$

$x + \frac{x^{2}}{2} - \frac{x^{3}}{3} - \ln | 1 + x | + C$

$x + \frac{x^{2}}{2} + \frac{x^{3}}{3} - \ln | 1 + x | + C$

answer is D.

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Detailed Solution

$\frac{x^{3}}{x + 1} = \frac{x^{3} + 1 - 1}{x + 1} = \frac{- 1}{(x + 1)} + (x^{2} + 1 - x)$

Thus given integral

$= \int (x^{2} + 1 - x - \frac{1}{1 + x}) d x = \frac{x^{3}}{3} + x - \frac{x^{2}}{2} - \ln | x + 1 | + C$

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