$\int \frac{\mathrm{d} x}{x\left(x^3+1\right)}=$

To evaluate $\int \frac{dx}{x(x^3+1)}$, begin by multiplying numerator and denominator by $x^2$:

$I = \int \frac{x^2 dx}{x^3(x^3+1)}$

Let $t = x^3$ so that $dt = 3x^2 dx$, hence $x^2 dx = \frac{1}{3} dt$.

Substituting gives:

$I = \frac{1}{3} \int \frac{dt}{t(t+1)}$

Decompose $\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$ by partial fractions.

Thus,

$I = \frac{1}{3} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt = \frac{1}{3} ( \ln|t| - \ln|t+1| ) + c = \frac{1}{3} \ln \left| \frac{t}{t+1} \right| + c$

Substitute $t = x^3$ back:

$I = \frac{1}{3} \ln \left| \frac{x^3}{x^3+1} \right| + c = \ln \left| \left( \frac{x^3}{x^3+1} \right)^{1/3} \right| + c$

This result corresponds to option C.

Final answer: $\boxed{\text{C}}$