# How do you integrate 2 / (x^3 + 1) using partial fractions?

Mar 14, 2018

Factorize the denominator then apply partial fraction decomposition.

#### Explanation:

Let

$I = \int \frac{2}{{x}^{3} + 1} \mathrm{dx}$

Factorize the denominator:

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

Apply partial fraction decomposition:

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

Factor out the term where the numerator is a multiple of the derivative of the denominator:

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

Complete the square in the denominator:

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

Integrate term by term:

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

Simplify:

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