Evaluate the integral $\int \frac{1}{\sqrt[3]{(x+1)^2(x-2)^2}}\mathrm dx$ What substitution is useful for this integral?
$$\int \frac{1}{\sqrt[3]{(x+1)^2(x-2)^2}}\mathrm dx$$
Substitutions $u=x^{\frac{2}{3}},u=(x+1)^{\frac{2}{3}},u=(x-2)^{\frac{2}{3}}$ are not working. 
Can't find useful trigonometric substitution. 

Change variable to 
$$x - \frac12 = \frac32 u \iff u = \frac{2x-1}{3}
\quad\text{ followed by }\quad
(u^2-1)^{1/3} = v \iff u = \sqrt{1+v^3}$$
We have
$$\int\frac{dx}{\sqrt[3]{(x+1)^2(x-2)^2}}
= \int\frac{dx}{(x^2-x-2)^{2/3}}
= \int\frac{dx}{\left(\left(x-\frac12\right)^2-\frac{9}{4}\right)^{2/3}}\\
= \left(\frac{2}{3}\right)^{1/3}\int\frac{du}{(u^2-1)^{2/3}}
= \left(\frac{2}{3}\right)^{1/3}\int\frac{d\sqrt{1+v^3}}{v^2}
= \left(\frac{3}{2}\right)^{2/3}\int \frac{dv}{\sqrt{1+v^3}}
$$
Using result from this answer,
the integral becomes
$$\frac{3^{5/12}}{2^{2/3}}
\left.F\left(\cos^{-1}\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\sqrt[3]{\frac49(x+1)(x-2)}}-1\right)\right|\frac{2+\sqrt{3}}{4}\right)
 + \text{constant}$$
where $\displaystyle\;F(\phi|m) = \int_0^\phi \frac{d\theta}{\sqrt{1-m(\sin\theta)^2}}\;$
is the incomplete elliptic integral of the first kind.
