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 :)
 A: Recognize
$$\frac{4x}{x^3-8}
= \frac{(x^2+2x+4) -x^2 +2(x-2)}{x^3-8}
= \frac1{x-2} - \frac{x^2}{x^3-8}+\frac2{x^2+2x+4}
$$
and integrate to obtain
$$4\int \frac x{x^3-8}dx=\ln|x-2|-\frac13\ln|x^3-8|+ \frac2{\sqrt3}\tan^{-1}\frac{x+1}{\sqrt3}
$$
A: hint
$$f(x)=\frac{x}{x^3-8}=$$
$$\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}$$
$$A=\frac{2}{12}=\frac 16$$
$$B=-A\;,\;C=2A$$
thus
$$f(x)=A(\frac{1}{x-2}-\frac 12\frac{2x+2-6}{x^2+2x+4})$$
and
$$\int f(x)dx=A(\ln(|x-2|)-\frac 12\ln(|x^2+2x+4|))+3A\int \frac{dx}{(x+1)^2+3}$$
A: 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*}
