$\displaystyle\int \dfrac{dx}{x^4 + 1}$.

$\dfrac{1}{x^4 + 1} = \dfrac{1}{(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)} =$

$= \dfrac{1}{4} \cdot \dfrac{-\sqrt{2}x + 2}{x^2 - \sqrt{2}x + 1} + \dfrac{1}{4} \cdot \dfrac{\sqrt{2}x + 2}{x^2 + \sqrt{2}x + 1}$

$\displaystyle\int \dfrac{dx}{x^4 + 1} = \dfrac{-1}{2} \displaystyle\int \dfrac{\sqrt{2}x - 2}{(\sqrt{2}x - 1)^2 + 1}\ dx\ + \dfrac{1}{2} \displaystyle\int \dfrac{\sqrt{2}x + 2}{(\sqrt{2}x + 1)^2 + 1}\ dx\ =$

 

$= \dfrac{-\sqrt{2}}{4} \displaystyle\int \dfrac{u - 1}{u^2 + 1}\ du + \dfrac{\sqrt{2}}{4} \displaystyle\int \dfrac{v + 1}{v^2 + 1}\ dv =$

$= \scriptsize{\fbox{$\dfrac{-\sqrt{2}}{8} \log [(\sqrt{2}x - 1)^2 + 1] + \dfrac{\sqrt{2}}{8} \log [(\sqrt{2}x + 1)^2 + 1] + \dfrac{\sqrt{2}}{4} \arctan (\sqrt{2}x - 1) + \dfrac{\sqrt{2}}{4} \arctan (\sqrt{2}x + 1) + c$}}$