# How do you integrate x/(1+x^4)?

$\int \frac{x}{1 + {x}^{4}} \mathrm{dx} = \frac{1}{2} \arctan \left({x}^{2}\right)$
$\int \frac{x}{1 + {x}^{4}} \mathrm{dx} = \frac{1}{2} \int \frac{2 x \mathrm{dx}}{1 + {\left({x}^{2}\right)}^{2}} = \frac{1}{2} \int \frac{d \left({x}^{2}\right)}{1 + {\left({x}^{2}\right)}^{2}} = \text{I}$
let ${x}^{2} = u$
$\implies \text{I} = \frac{1}{2} \int \frac{\mathrm{du}}{1 + {u}^{2}} = \frac{1}{2} \arctan u = \frac{1}{2} \arctan \left({x}^{2}\right)$