Evaluate: ∫ (x(1 + x²))/(1 + x⁴) dx

$\frac{x(1+x^2)}{1+x^4} = \frac{x}{1+x^4}+\frac{x^3}{1+x^4}$ For the first term put $x^2=t$, $x\,dx=\frac{dt}{2}$: $\int\frac{x\,dx}{1+x^4} = \frac12\int\frac{dt}{1+t^2} = \frac12\tan^{-1}x^2$. For the second term put $1+x^4=u$, $x^3dx=\frac{du}{4}$: $\int\frac{x^3\,dx}{1+x^4} = \frac14\log|u| = \frac14\log(1+x^4)$. Hence the integral is $\frac12\tan^{-1}(x^2)+\frac14\log(1+x^4)+C$.