Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$ I need to solve the following integrals:
$$\int{\frac{1}{(x^3+1)^2}}dx$$
and
$$\int{\frac{1}{(x^3-1)^2}}dx$$
My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?

The standard method uses partial fractions decomposition.
Here is how it begins for the first integral: factorout  the denominator into irreducible factors with high-school identities:
$$x^3+1=(x+1)(x^2-x+1)$$
Newt proceed to decompose into partial fractions:
$$\frac{1}{(x^3+1)^2}=\frac{1}{(x+1)^2(x^2-x+1)^2}=\frac A{x+1}+\frac B{(x+1)^2}+\frac {Cx+D}{x^2-x+1}+\frac{Ex+F}{(x^2-x+1)^2}.$$
