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Hola csanchez!
$$\begin{align}&\int \frac 1 {x(x^3-2)}dx=\\&\\&u=x^3\ ==> du=3x^2dx\\&\\&dx= \frac {du}{3x^2}\ ===>\frac{dx} x=\frac{du}{3x^3}= \frac {du} {3u}\\&\\&\int \frac 1 {u-2} \frac {du} {3u}= \frac 1 3 \int \frac 1 {u^2-2u}du=\\&\\&completando \ cuadrados:\\&\\&= \frac 1 3 \int \frac 1 {(u-1)^2-1} du\\&\\&u-1=t\ ==>du=dt\\&\\&= \frac 1 3 \int \frac 1 {t^2-1}dt=- \frac 1 3 \int \frac 1 {1-t^2} dt=\\&\\&=-\frac 1 3 arctanht=- \frac 1 3 arctanh(u-1)= - \frac 1 3 arctanh(x^3-1)=\\&\\&=\frac 1 3 arc tanh(1-x^3)+C\end{align}$$
Saludos y recuerda votar
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