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

Answer 2

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

Answer 3

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*}