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

Mar 23, 2017

Let's first determine a factorisation for ${x}^{3} + {x}^{2} - x - 1$.

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

The partial fraction decomposition will be of the form

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

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

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

$A {x}^{2} + 2 A x + A + B {x}^{2} + C x - B x - C = 3 x - 2$

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

We now get the system of equations $\left\{\begin{matrix}A + B = 0 \\ 2 A - B + C = 3 \\ A - C = - 2\end{matrix}\right.$

Solve to get

$2 A - \left(- A\right) + A + 2 = 3$

$A = \frac{1}{4}$

$B = - \frac{1}{4}$

$C = \frac{9}{4}$

The integral becomes

$\int \frac{1}{4 \left(x - 1\right)} - \frac{\frac{1}{4} x - \frac{9}{4}}{x + 1} ^ 2 \mathrm{dx}$

$\int \frac{5}{4 \left(x - 1\right)} \mathrm{dx} - \int \frac{x - 9}{4 {\left(x + 1\right)}^{2}} \mathrm{dx}$

We can rewrite $x - 9$ as $\left(x + 1\right) - 10$. Therefore, the second integral is equivalent to

$- \frac{1}{4} \left(\int \frac{1}{x + 1} - \frac{10}{x + 1} ^ 2\right) \mathrm{dx}$

This can be integrated as

$- \frac{1}{4} \ln | x + 1 | - \frac{5}{2 \left(x + 1\right)}$

The other part of the integral is equal to $\frac{1}{4} \ln | x - 1 |$. Putting all of this together, we have

$\frac{1}{4} \ln | x - 1 | - \frac{1}{4} \ln | x + 1 | - \frac{5}{2 \left(x + 1\right)} + C$

By laws of logarithms, we have

$\frac{1}{4} \ln | \frac{x - 1}{x + 1} | - \frac{5}{2 \left(x + 1\right)} + C$

Hopefully this helps!