Integrate $\int\frac{x}{x^3-8}dx$

Question

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 🙂

Answer 1

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 $$4\int \frac x{x^3-8}dx=\ln|x-2|-\frac13\ln|x^3-8|+ \frac2{\sqrt3}\tan^{-1}\frac{x+1}{\sqrt3} $$