Integral $\int\frac{dx}{(x^3-1)^2}$

Question

Please help. I do not know what to do. You can just show the direction where to go and I continue. Here it is: $$\int\frac{dx}{(x^3-1)^2}$$

Answer 1

Hint: Rewrite your integrand as $$\frac 1{(x^3-1)^2}=\frac 1{(x-1)^2(x^2+x+1)^2}$$ and consider doing partial fractions on the latter.

Answer 2

Let $$\displaystyle I = \int\frac{1}{(x^3-1)^2}dx = \frac{1}{3}\int\frac{1}{x^2}\cdot \frac{3x^2}{(x^3-1)^2}dx$$

Now Using Integration by parts, we get

$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{1}{(x^3-1)}-\frac{2}{3}\int \left[\frac{1}{x^3\cdot (x^3-1)}\right]dx$$

$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{1}{(x^3-1)}+\frac{2}{3}\int \left[\frac{1}{x^3}-\frac{1}{(x^3-1)}\right]dx$$

$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{1}{(x^3-1)}+\frac{2}{3}\int x^{-3}dx-\frac{2}{3}\int\frac{1}{(x-1)(x^2+x+1)}dx$$

$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{1}{(x^3-1)}-\frac{1}{3x^2}-\frac{2}{3}\int\frac{1}{(x-1)(x^2+x+1)}dx$$

Now Using partial fraction descomposition

$$\displaystyle \frac{1}{(x-1)(x^2+x+1)} = \frac{A}{x-1}+\frac{Bx+c}{x^2+x+1}$$

Now camparing Coefficients, we get

$$\displaystyle A=\frac{1}{3}\;, B=-\frac{1}{3}\;,C=-\frac{2}{3}$$

So $$\displaystyle \int\frac{1}{(x-1)(x^2+x+1)} = \frac{1}{3}\int\frac{1}{x-1}-\frac{1}{3}\int\frac{x+2}{x^2+x+1}dx$$

$$\displaystyle =\frac{1}{3}\int\frac{1}{x-1}-\frac{1}{6}\int\frac{2x+1}{x^2+x+1}-\frac{1}{2}\int\frac{1}{x^2+x+1}dx$$

$$\displaystyle = \frac{1}{3}\ln |x-1|-\frac{1}{6}\ln|x^2+x+1|-\frac{1}{2}\int\frac{1}{x^2+x+1}dx$$