# How do you use partial fractions to find the integral int (4x^2)/(x^3+x^2-x-1)dx?

Dec 5, 2016

$\int \frac{4 {x}^{2} \mathrm{dx}}{{x}^{3} + {x}^{2} - x - 1} = \ln | x - 1 | + 3 \ln | x + 1 | + \frac{2}{x + 1}$

#### Explanation:

The general method to integrate proper rational functions is to develop them in a sum of partial fractions.

In order to do that, we must first factorize the denominator.
You can easily see (x+1) and (x-1) are factors and write:

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

The decomposition in partial fraction is thus:

$\frac{4 {x}^{2}}{{x}^{3} + {x}^{2} - x - 1} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{{\left(x + 1\right)}^{2}}$

$\frac{4 {x}^{2}}{{\left(x + 1\right)}^{2} \left(x - 1\right)} = \frac{A {\left(x + 1\right)}^{2} + B \left(x - 1\right) \left(x + 1\right) + C \left(x - 1\right)}{{\left(x + 1\right)}^{2} \left(x - 1\right)}$

$A {x}^{2} + 2 A x + A + B {x}^{2} - B + C x - C = 4 {x}^{2}$

Equating the coefficientS of the same grade in $x$ we have the linear system:

$A + B = 4$
$2 A + C = 0$
$A - B - C = 0$

that can be solved as:

$A = 1 , B = 3 , C = - 2$

So:

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

$\int \frac{4 {x}^{2} \mathrm{dx}}{{x}^{3} + {x}^{2} - x - 1} = \ln | x - 1 | + 3 \ln | x + 1 | + \frac{2}{x + 1}$