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$\int \frac{x^{4}}{x^{2} + 1} d x$ is equal to

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$\frac{x^{3}}{3} + x + \tan^{- 1} x + C$

$\frac{x^{3}}{3} - x + \tan^{- 1} + C$

$\frac{x^{2}}{2} - x + 2 \tan^{- 1} x + C$

$\frac{x^{3}}{3} - x - \tan^{- 1} x + C$

answer is B.

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Detailed Solution

$Since degree of numerator is greater than the$

$degree of denominator, we divide them$

$\frac{x^{4}}{x^{2} + 1} = x^{2} - 1 + \frac{1}{x^{2} + 1}$

so $\int \frac{x^{4}}{x^{2} + 1} d x = \frac{x^{3}}{3} - x + \tan^{- 1} x + C$

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