Evaluate:
$$\int \dfrac{2\pi y}{2y^3-1}\,dy$$
I've been struggling with this for a while. If it had just been $y^3$ instead of $2y^3$ in the Denominator, Partial Fraction Decomposition, although quite long, would be one method of evaluating the Integral. However, I just cannot understand what to do with this Integral.$$$$I would be indeed very grateful for any help in solving this Integral. Many thanks in advance!
Use partial fraction decomposition. Express
$$\frac{x}{x^3-1} = \frac{A x+B}{x^2+x+1}+\frac{C}{x-1} $$
where $x=2^{1/3} y$. Note that
$$x^2+x+1 = \left ( x+\frac12 \right )^2+\frac34$$
$$A x+B = A \left ( x+\frac12 \right ) + B - \frac12 A$$
$$\int \frac{dx}{\left ( x+\frac12 \right )^2+\frac34} = \frac{2}{\sqrt{3}} \arctan{\left (\frac{2 x+1}{\sqrt{3}}\right )}$$
$$\int dx \frac{x+\frac12}{\left ( x+\frac12 \right )^2+\frac34} = \frac12 \log{(x^2+x+1)}$$