Can the integral $\int \frac{6x-1}{3x^3-4} \, dx$ solved with a simple solution? I got the integral 
$$\int \frac{6x-1}{3x^3-4} \, dx$$
as an exercise for exam preparation. I tried to get closer to the result with third binomial formula, but could not get anywhere. So I tried to find the solution with wolframalpha, but this one seems overly complex. Is there a simple solution for this integral?
 A: $$\int \frac{6x-1}{3x^3-4} \, dx$$
$$\frac{1}{4}\int \frac{6x-1}{\frac{3}{4}x^3-1} \, dx$$
$\frac{3}{4}x^3=u^3$
$\sqrt[3]{\frac{3}{4}}x=u$
$\sqrt[3]{\frac{4}{3}}u=x$
$\sqrt[3]{\frac{4}{3}}du=dx$
$$\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{6\sqrt[3]{\frac{4}{3}}u-1}{u^3-1} \, du=\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{A}{u-1} +\frac{B}{u-\epsilon}+\frac{C}{u-\epsilon^2}\, du$$
where $\epsilon=e^{\frac{2\pi i}{3} }=-\frac{1}{2}+i\frac{\sqrt3}{2}$
$\epsilon^2=e^{\frac{4\pi i}{3} }=-\frac{1}{2}-i\frac{\sqrt3}{2}$
$$\frac{6\sqrt[3]{\frac{4}{3}}u-1}{u^3-1}=\frac{A}{u-1} +\frac{B}{u-\epsilon}+\frac{C}{u-\epsilon^2}$$
$f(u)=6\sqrt[3]{\frac{4}{3}}u-1$
$A=f(1)=6\sqrt[3]{\frac{4}{3}}-1$
$B=f(\epsilon)=6\epsilon\sqrt[3]{\frac{4}{3}}-1$
$C=f(\epsilon^2)=6\epsilon^2\sqrt[3]{\frac{4}{3}}-1$
$$\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{A}{u-1} +\frac{B}{u-\epsilon}+\frac{C}{u-\epsilon^2}\, du=$$
$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-1} \, du+\frac{B\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon}\, du+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon^2}\, du=$$
$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-1} \, du+\frac{B\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon}\, du+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon^2}\, du=$$
$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\ln {(u-1)} +\frac{B\sqrt[3]{\frac{4}{3}}}{4}\ln {(u-\epsilon)}+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\ln {(u-\epsilon^2)}+c=$$
$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\ln {(\sqrt[3]{\frac{3}{4}}x-1)} +\frac{B\sqrt[3]{\frac{4}{3}}}{4}\ln {(\sqrt[3]{\frac{3}{4}}x-\epsilon)}+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\ln {(\sqrt[3]{\frac{3}{4}}x-\epsilon^2)}+c=$$
After that you will need to do some calculations and also  need to do transform of the complex value of $\ln$ couple  to $arctan$
Use the formula for that transform  $\arctan x = \frac{1}{2}i\left(\ln\left(1-i\,x\right)-\ln\left(1+i\,x\right)\right) $
Ref :http://en.wikipedia.org/wiki/Inverse_trigonometric_functions
The most easiest way for me is that way. Maybe someone else can offer quicker way.
A: Use partial fractions. Factor the denominator as
$$3x^3-4=3(x^3-4/3)=3(x-(4/3)^{1/3})(x^2+(4/3)^{1/3}x+(4/3)^{2/3}),$$
then set up with $A$ over the linear factor, and $Bx+C$ over the quadratic factor.
The answer is messy, so the values of $A,B,C$ will be complicated. However the integral will be doable using $\ln(u),\arctan(u)$ as usual with partial fractions technique.
