Answer
$$\frac{1}{4}\left(4\ln \left(x\right)-\ln \left|1+x^4\right|\right)+C$$
Work Step by Step
Given
$$\int \frac{d x}{x\left(x^4+1\right)}$$
Let
$$ u=x^4+1\ \ \ \to\ \ \ du = 4x^3 dx$$
Then
\begin{aligned} \int \frac{d x}{x\left(x^4+1\right)}&= \int \frac{du}{4x^4\left(x^4+1\right)}\\
&=\int \frac{d u}{4(u-1)u}\end{aligned}
Usin partila fraction, since
\begin{aligned}
\frac{1}{u\left(u-1\right)}&=\frac{a_0}{u}+\frac{a_1}{u-1}\\
&= \frac{a_0\left(u-1\right)+a_1u}{u(u-1)}\\
1&=a_0\left(u-1\right)+a_1u
\end{aligned}
at $u=0\to a_0 = -1,\ \ \ u=1 \ \ \to a_1= 1$, then
\begin{aligned}
\frac{1}{\left(u\right)\left(u-1\right)}= -\frac{1}{u}+\frac{1}{u-1}
\end{aligned}
Hence
\begin{aligned} \int \frac{d x}{x\left(x^4+1\right)}&= \int \frac{du}{4x^4\left(x^4+1\right)}\\
&=\int \frac{d u}{4(u-1)u}\\
&= \frac{1}{4}\left(\int \frac{1}{u-1}du-\int \frac{1}{u}du\right)\\
&= \frac{1}{4}\left(\ln \left|u-1\right|-\ln \left|u\right|\right)\\
&= \frac{1}{4}\left(4\ln \left(x\right)-\ln \left|1+x^4\right|\right)+C\end{aligned}
