Integrate the rational function: $\frac{1}{x(x^{4}-1)}$

(N/A) To integrate $\int \frac{1}{x(x^{4}-1)} dx$,multiply the numerator and denominator by $x^{3}$:
$\int \frac{x^{3}}{x^{4}(x^{4}-1)} dx$
Let $x^{4} = t$,then $4x^{3} dx = dt$,which implies $x^{3} dx = \frac{1}{4} dt$.
Substituting these into the integral:
$\frac{1}{4} \int \frac{dt}{t(t-1)}$
Using partial fractions,$\frac{1}{t(t-1)} = \frac{A}{t} + \frac{B}{t-1}$.
$1 = A(t-1) + Bt$. Setting $t=0$,we get $A=-1$. Setting $t=1$,we get $B=1$.
Thus,$\frac{1}{4} \int (\frac{-1}{t} + \frac{1}{t-1}) dt = \frac{1}{4} [-\ln|t| + \ln|t-1|] + C$
$= \frac{1}{4} \ln|\frac{t-1}{t}| + C$
$= \frac{1}{4} \ln|\frac{x^{4}-1}{x^{4}}| + C$,where $C$ is the constant of integration.