Integrate $\int\frac{x}{x^3-8}dx$ Integrate
$$\int\frac{x}{x^3-8}dx$$
I solved this integral by dividing it in partial fractions, than i came in two integrals $I_1,I_2$. Then, partioning $I_2$ into $I_3,I_4$ and partioning $I_4$ into $I_5,I_6$. But it took me so much work even though i got correct answer.
Is there any simple way to solve it?
Thank you in advance :)

Partial fractions will indeed work without the need for too many integrals. We see that
\begin{align*}
\int \frac{x}{x^3 -8} \ dx &= \int \frac{1}{6}\frac{1}{x-2} - \frac{1}{6} \frac{x-2}{x^2 +2x+4}\ dx \\
&= \frac{1}{6} \ln(x-2) - \frac{1}{6} \int  \frac{\frac{1}{2}(2x+2)-3}{x^2 +2x+4}\ dx\\
&\overset{\color{blue}{u=x^2 +2x+4}}= \frac{1}{6} \ln(x-2) - \frac{1}{12} \int  \frac{1}{u}\ du + \frac{1}{2} \int \frac{1}{(x+1)^2 +3} \ dx\\
&\overset{\color{blue}{s=x+1}}= \frac{1}{6} \ln(x-2) - \frac{1}{12} \ln(x^2 +2x+4)+ \frac{1}{2} \int \frac{1}{s^2 +(\sqrt{3})^2} \ ds\\
&= \frac{1}{6} \ln(x-2) - \frac{1}{12} \ln(x^2 +2x+4)+ \frac{1}{2} \frac{\arctan\left(\frac{x+1}{\sqrt{3}} \right)}{\sqrt{3}} + C
\end{align*}
