Step 1: Plan a smart move.
The integral $\int\frac{dx}{x(x^4 + 1)}$ is hard as it stands. The trick is to multiply top and bottom by $x^3$, which sets up a clean substitution.
Step 2: Multiply by $x^3$.
Multiplying top and bottom by $x^3$ turns the bottom's single $x$ into $x^4$.
\[ I = \int\frac{x^3}{x^4(x^4 + 1)}\,dx \]
Step 3: Substitute $u = x^4$.
Then $du = 4x^3\,dx$, so $x^3\,dx = \frac{du}{4}$. The integral becomes simpler.
\[ I = \frac{1}{4}\int\frac{du}{u(u + 1)} \]
Step 4: Split using partial fractions.
We can write $\frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1}$.
\[ I = \frac{1}{4}\int\left(\frac{1}{u} - \frac{1}{u+1}\right)du \]
Step 5: Integrate each piece.
Each gives a logarithm. Combine them using the log subtraction rule.
\[ I = \frac{1}{4}\left(\ln|u| - \ln|u+1|\right) + C = \frac{1}{4}\ln\left|\frac{u}{u+1}\right| + C \]
Step 6: Put $u = x^4$ back.
Replace $u$ with $x^4$.
\[ I = \frac{1}{4}\ln\left|\frac{x^4}{x^4 + 1}\right| + C \]
\[ \boxed{\tfrac{1}{4}\ln\left|\tfrac{x^4}{x^4 + 1}\right| + C} \]