∫ (x^6 1x^2 + 1) dx=∫x^6+1 2x^2+1=∫x^6+1x^2+1 ∫21+x^2=∫(x^2+1)(x^4+1 x^2)x^2+1 2tan^ 1x+c =∫ (x^4 x^2+1) 2tan^ 1x+c=x^55 x^33+x 2tan^ 1x+c ...

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Integration by Partial Fractions

∫ x6 -1/x2 + ...

Question

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

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Solution

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

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