I=∫11+x4dx=12∫(x2+1)−(x2−1)x4+1dx=12∫x2+1x4+1dx−12∫x2−1x4+1dx =12∫1+1x2x-1x2+22dx−12∫1−1x2x+1x2-22dx put x-1x=t and x+1x=t then=122Tan−1x2−12x−142logx2−2x+1x2+2x+1+c

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$\int \frac{1}{x^{4} + 1} d x =$

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$\frac{1}{2 \sqrt{2}} {Tan}^{- 1} (\frac{x^{2} + 1}{\sqrt{2} x}) - \frac{1}{4 \sqrt{2}} \log |\frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1}| + c$

$\frac{1}{2 \sqrt{2}} {Tan}^{- 1} (\frac{x^{2} - 1}{\sqrt{2} x}) - \frac{1}{4 \sqrt{2}} \log |\frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1}| + c$

$\frac{1}{\sqrt{2}} {Tan}^{- 1} (\frac{x^{2} - 1}{\sqrt{2} x}) - \frac{1}{2 \sqrt{2}} \log |\frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1}| + c$

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Detailed Solution

$I = \int \frac{1}{1 + x^{4}} d x = \frac{1}{2} \int \frac{(x^{2} + 1) - (x^{2} - 1)}{x^{4} + 1} d x = \frac{1}{2} \int \frac{x^{2} + 1}{x^{4} + 1} d x - \frac{1}{2} \int \frac{x^{2} - 1}{x^{4} + 1} d x = \frac{1}{2} \int \frac{(1 + \frac{1}{x^{2}})}{{(x - \frac{1}{x})}^{2} + {(\sqrt{2})}^{2}} d x - \frac{1}{2} \int \frac{(1 - \frac{1}{x^{2}})}{{(x + \frac{1}{x})}^{2} - {(\sqrt{2})}^{2}} d x p u t x - \frac{1}{x} = t a n d x + \frac{1}{x} = t t h e n$

$= \frac{1}{2 \sqrt{2}} {Tan}^{- 1} (\frac{x^{2} - 1}{\sqrt{2} x}) - \frac{1}{4 \sqrt{2}} \log |\frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1}| + c$