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integrate [1/((x)^4+1)]dx

$I = \int \frac{1}{x^{4} + 1} ⅆ x = \int \frac{\frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x = \frac{1}{2} \int \frac{\frac{2}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x = \frac{1}{2} \int (\frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} - \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}}) d x = \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x - \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x = \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 2} d x - \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 2} d x p u t x - \frac{1}{x} = u i n f i r s t int e g r a l & x + \frac{1}{x} = v i n \sec o n d int e g r a l = \frac{1}{2} \int \frac{1}{u^{2} + {(\sqrt{2})}^{2}} d u - \frac{1}{2} \int \frac{1}{v^{2} - {(\sqrt{2})}^{2}} d v = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) - \frac{1}{2} * \frac{1}{2 \sqrt{2}} \log | \frac{v - \sqrt{2}}{v + \sqrt{2}} | + C = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{2} x}) - \frac{1}{4 \sqrt{2}} \log | \frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1} | + C$

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