Answer
$-\frac{1}{x^{4}+1}$+C
Work Step by Step
Put u= $x^{4}+1$.
$\frac{du}{dx}= 4x^{3}$ or dx=$\frac{du}{4x^{3}}$
Then, I= $\int(\frac{4x^{3}}{u^{2}}\times\frac{du}{4x^{3}})=\int u^{-2}du$
=$ \frac{u^{-2+1}}{-2+1}+C= \frac{-1}{u}+C$.
Substituting the value of u, we have
$\int \frac{4x^{3}}{(x^{4}+1)^{2}}dx=-\frac{1}{x^{4}+1}$+C
