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 :)

  • 1
    $\begingroup$ Can you display what your $I_n$'s are? $\endgroup$
    – Xoque55
    Apr 17, 2021 at 22:02

3 Answers 3



$$\frac{4x}{x^3-8} = \frac{(x^2+2x+4) -x^2 +2(x-2)}{x^3-8} = \frac1{x-2} - \frac{x^2}{x^3-8}+\frac2{x^2+2x+4} $$ and integrate to obtain $$4\int \frac x{x^3-8}dx=\ln|x-2|-\frac13\ln|x^3-8|+ \frac2{\sqrt3}\tan^{-1}\frac{x+1}{\sqrt3} $$

  • 2
    $\begingroup$ you are completely another level bro always give me special edition approaches,thank you so much and have a good night :) $\endgroup$
    – Zenit
    Apr 17, 2021 at 22:58
  • $\begingroup$ @John - my pleasure ... $\endgroup$
    – Quanto
    Apr 17, 2021 at 23:26
  • $\begingroup$ @John I agree with you, Quanto is one of the best integral solvers on the site. You can accept his answer if it was the most helpful. $\endgroup$ Apr 20, 2021 at 19:32
  • $\begingroup$ How did you even manage to split the numerator in such a way? $\endgroup$ Nov 21 at 13:35


$$f(x)=\frac{x}{x^3-8}=$$ $$\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}$$

$$A=\frac{2}{12}=\frac 16$$


thus $$f(x)=A(\frac{1}{x-2}-\frac 12\frac{2x+2-6}{x^2+2x+4})$$

and $$\int f(x)dx=A(\ln(|x-2|)-\frac 12\ln(|x^2+2x+4|))+3A\int \frac{dx}{(x+1)^2+3}$$

  • 3
    $\begingroup$ i am assuming op doesn’t want another partial fractions method; this seems like it leads to the same solution that the op wanted to avoid $\endgroup$
    – C Squared
    Apr 17, 2021 at 22:06
  • $\begingroup$ i see now. sorry about that $\endgroup$
    – C Squared
    Apr 17, 2021 at 22:10
  • $\begingroup$ @hamam my $I_4$ is this $3A\cdot\int\frac{x}{(x+1)^2+3}dx\:\:$ so i did this$\:\:\frac{1}{2}\int\frac{2x-2+2}{(x+1)^2)+3}dx$ and evaluated in $I_4,I_5$ $\endgroup$
    – Zenit
    Apr 17, 2021 at 22:18
  • $\begingroup$ $I_6=\frac{1}{6\sqrt{3}}\cdot\arctan(\frac{x+1}{\sqrt{3}})+C$ cause i carried coefficients before Integral from beginning $\endgroup$
    – Zenit
    Apr 17, 2021 at 22:23
  • $\begingroup$ @hamam just added all my $I_n$ and i have same result as you just some coefficients prob. that noticed, now im tryin to shorten my steps as your evaluation thank you :) $\endgroup$
    – Zenit
    Apr 17, 2021 at 22:52

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*}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .