Answer
$$\int\frac{1}{x(x^4+1)}dx=\frac{1}{4}\ln\frac{x^4}{x^4+1}+C$$
Work Step by Step
$$I=\int\frac{1}{x(x^4+1)}dx$$
Multiply both numerator and denominator with $x^3$:
$$I=\int\frac{x^3}{x^4(x^4+1)}dx$$
We set $u=x^4$, so $$du=4x^3dx$$ $$x^3dx=\frac{1}{4}du$$
That means
$$I=\frac{1}{4}\int\frac{1}{u(u+1)}du$$ $$I=\frac{1}{4}\int\frac{(u+1)-u}{u(u+1)}du$$ $$I=\frac{1}{4}\Big(\int\frac{du}{u}-\int\frac{du}{u+1}\Big)$$ $$I=\frac{1}{4}(\ln|u|-\ln|u+1|)+C$$ $$I=\frac{1}{4}\ln|\frac{u}{u+1}|+C$$ $$I=\frac{1}{4}\ln|\frac{x^4}{x^4+1}|+C$$
Since $x^4\ge0$ for all $x$, $|\frac{x^4}{x^4+1}|=\frac{x^4}{x^4+1}$
$$I=\frac{1}{4}\ln\frac{x^4}{x^4+1}+C$$
