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

Oct 31, 2016

The integral is $\frac{1}{3} \ln \left(x - 1\right) - \frac{1}{6} \ln \left({x}^{2} + x + 1\right) - \frac{1}{\sqrt{3}} \arctan \left(\frac{2 x + 1}{\sqrt{3}}\right)$

#### Explanation:

Let's factorise the denominator
${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$
and the decomposition in partial fractions
$\frac{1}{{x}^{3} - 1} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + x + 1}$
$1 = A \left({x}^{2} + x + 1\right) + \left(B x + C\right) \left(x - 1\right)$
If $x = 0$$\implies$$1 = A - C$
coefficients of x^2, $0 = A + B$
Coefficients of x, $0 = A - B + C$
Solving, we found $A = \frac{1}{3} , B = - \frac{1}{3} , C = - \frac{2}{3}$
so $\int \frac{\mathrm{dx}}{{x}^{3} - 1} = \frac{1}{3} \int \frac{\mathrm{dx}}{x - 1} - \frac{1}{3} \int \frac{\left(x + 2\right) \mathrm{dx}}{{x}^{2} + x + 1}$
$\int \frac{\mathrm{dx}}{x - 1} = \ln \left(x - 1\right)$
$\int \frac{\left(x + 2\right) \mathrm{dx}}{{x}^{2} + x + 1} = \frac{1}{2} \int \frac{\left(2 x + 1\right) \mathrm{dx}}{{x}^{2} + x + 1} + \frac{3}{2} \int \frac{\mathrm{dx}}{{x}^{2} + x + 1}$
$\frac{1}{2} \int \frac{\left(2 x + 1\right) \mathrm{dx}}{{x}^{2} + x + 1} = \frac{1}{2} \ln \left({x}^{2} + x + 1\right)$
$\int \frac{\mathrm{dx}}{{x}^{2} + x + 1} = \int \frac{\mathrm{dx}}{{\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4}}$
let's do the substitution,$u = \frac{2 x + 1}{\sqrt{3}}$$\implies$$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{2}{\sqrt{3}}$
$\int \frac{\mathrm{dx}}{{\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4}} = \frac{2}{\sqrt{3}} \int \frac{\mathrm{du}}{{u}^{2} + 1} = \frac{2}{\sqrt{3}} \arctan u$