$\int \frac{x^{2} + 1}{x (x^{2} - 1)} d x$ is equal to

log \frac{x^{2} - 1}{x} + C

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- log \frac{x^{2} - 1}{x} + C

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log \frac{x}{x^{2} + 1} + C

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- log \frac{x}{x^{2} + 1} + C

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Solution

The correct option is B $log \frac{x^{2} - 1}{x} + C$
$\int \frac{x^{2} + 1}{x (x^{2} - 1)} d x$
Take partial fraction
$\int \frac{x^{2} + 1}{x . (x^{2} - 1)} d x = \int \frac{- 1}{x} + \frac{1}{x + 1} + \frac{1}{x - 1} d x$
$= - l n | x | + l n (x + 1) + l n (x - 1) + c$ .

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