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

Jul 11, 2017

$\int \frac{1}{{x}^{3} - 1} \mathrm{dx} = \frac{1}{6} \left[2 \ln | x - 1 | - \ln | {x}^{2} + x + 1 | - 2 \sqrt{3} \arctan \left(\frac{\left(2 x + 1\right) \sqrt{3}}{3}\right)\right] + \text{c}$

#### Explanation:

We need to find the partial fraction expansion of $\frac{1}{{x}^{3} - 1}$.

Firstly, we need to rewrite ${x}^{3} - 1$ as a product of irreducible polynomials. ${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$. It's easy to factorise this function since we know $\left(x - 1\right)$ is a root. Then we can use inspection or polynomial division to find the other function.

$\therefore \frac{1}{{x}^{3} - 1} = \frac{1}{\left(x - 1\right) \left({x}^{2} + x + 1\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + x + 1}$

Since the numerator has to have a polynomial to a power one less than the denominator, the second term will have to have an $x$ in the numerator.

$1 = A \left({x}^{2} + x + 1\right) + \left(B x + C\right) \left(x - 1\right)$

$A + B = 0 \Rightarrow A = - B$ $\left(1\right)$

$A - B + C = 0$ $\left(3\right)$

$A - C = 1 \Rightarrow C = A - 1$ $\left(2\right)$

Sub $\left(1\right)$ and $\left(2\right)$ into $\left(3\right)$

$A + A + A - 1 = 0$

$3 A = 1$

$A = \frac{1}{3}$

$B = - \frac{1}{3}$

$C = \frac{1}{3} - 1 = - \frac{2}{3}$

$\therefore \frac{1}{\left(x - 1\right) \left({x}^{2} + x + 1\right)} = \frac{1}{3 \left(x - 1\right)} + \frac{- \frac{1}{3} x - \frac{2}{3}}{{x}^{2} + x + 1}$

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

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

$= \frac{1}{3} \ln | x - 1 | + {\text{c}}_{1} - \frac{1}{3} \int \frac{x + 2}{{x}^{2} + x + 1}$ $\mathrm{dx}$

(x+2)/(x^2+x+1) = 1/2((2(x+2))/(x^2+x+1))

$= \frac{1}{2} \left(\frac{2 x + 1}{{x}^{2} + x + 1} + \frac{3}{{x}^{2} + x + 1}\right)$

$\therefore - \frac{1}{3} \int \frac{x + 2}{{x}^{2} + x + 1}$ $\mathrm{dx}$

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

$= - \frac{1}{6} \ln | {x}^{2} + x + 1 | + {\text{c}}_{2} - \frac{1}{6} \int \frac{3}{{x}^{2} + x + 1}$ $\mathrm{dx}$

$- \frac{1}{6} \int \frac{3}{{x}^{2} + x + 1}$ $\mathrm{dx} = - \frac{3}{6} \int \frac{1}{{\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4}}$ $\mathrm{dx}$

Let $s = x + \frac{1}{2}$ and $\mathrm{ds} = \mathrm{dx}$

$- \frac{1}{2} \int \frac{1}{{\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4}}$ $\mathrm{dx} = - \frac{1}{2} \int \frac{1}{{s}^{2} + \frac{3}{4}}$ $\mathrm{ds}$

$= - \frac{1}{2} \int \frac{\frac{4}{3}}{\frac{4}{3} \left({s}^{2} + \frac{3}{4}\right)}$ $\mathrm{ds} = - \frac{2}{3} \int \frac{1}{\frac{4}{3} {s}^{2} + 1}$ $\mathrm{ds}$

Let $p = \frac{2}{\sqrt{3}} s$ and $\mathrm{dp} = \frac{2}{\sqrt{3}} \mathrm{ds}$

$- \frac{2}{3} \int \frac{1}{\frac{4}{3} {s}^{2} + 1}$ $\mathrm{ds} = - \frac{1}{\sqrt{3}} \int \frac{1}{{p}^{2} + 1}$ $\mathrm{dp}$

$= - \frac{1}{\sqrt{3}} \arctan p + {\text{c"_3=-sqrt3/3arctan(((2x+1)sqrt3)/3)+"c}}_{3}$

Combining all three integrals gives

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

$- \frac{\sqrt{3}}{3} \arctan \left(\frac{\left(2 x + 1\right) \sqrt{3}}{3}\right) + \text{c}$

$= \frac{1}{6} \left[2 \ln | x - 1 | - \ln | {x}^{2} + x + 1 | - 2 \sqrt{3} \arctan \left(\frac{\left(2 x + 1\right) \sqrt{3}}{3}\right)\right] + \text{c}$