How do you integrate int 1/(x^3 -1) using partial fractions?

May 22, 2018

Separate into integrable parts.

Explanation:

Let

$I = \int \frac{1}{{x}^{3} - 1} \mathrm{dx}$

Factorize:

$I = \int \frac{1}{\left(x - 1\right) \left({x}^{2} + x + 1\right)} \mathrm{dx}$

Apply partial fraction decomposition:

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

Rearrange:

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

Complete the square in the last term:

$I = \frac{1}{3} \int \left(\frac{1}{x - 1} - \frac{1}{2} \frac{2 x + 1}{{x}^{2} + x + 1} - \frac{6}{{\left(2 x + 1\right)}^{2} + 3}\right) \mathrm{dx}$

Integrate term by term:

$I = \frac{1}{3} \left\{\ln | x - 1 | - \frac{1}{2} \ln | {x}^{2} + x + 1 | + \sqrt{3} {\tan}^{- 1} \left(\frac{2 x + 1}{\sqrt{3}}\right)\right\} + C$

Simplify:

$I = \frac{1}{3} \ln | x - 1 | - \frac{1}{6} \ln | {x}^{2} + x + 1 | + \frac{1}{\sqrt{3}} {\tan}^{- 1} \left(\frac{2 x + 1}{\sqrt{3}}\right) + C$