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Functions

Solve ∫x6 - ...

Question

Solve
$\int \frac{x^{6} - 1}{1 + x^{2}} d x$

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Solution

$x^{6} - 1 = (x^{2} + 1) (x^{4} - x^{2} + 1) \frac{x^{6} - 1}{x^{2} + 1} = x^{4} - x^{2} + 1 - \frac{2}{x^{2} + 1} \int \frac{x^{6} - 1}{x^{2} + 1} = \int x^{4} - x^{2} + 1 - 2 \int \frac{1}{x^{2} + 1} = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 {tan}^{- 1} (x) + C$

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