# How do you integrate (x^3-2x^2-4)/(x^3-2x^2) using partial fractions?

Oct 29, 2016

THe answer is $= x - \frac{2}{x} + \ln x - \ln \left(x - 2\right) + C$

#### Explanation:

Let's start by rewriting the expression
$\frac{{x}^{3} - 2 {x}^{2} - 4}{{x}^{3} - 2 {x}^{2}} = 1 - \frac{4}{{x}^{3} - 2 {x}^{2}} = 1 - \frac{4}{\left({x}^{2}\right) \left(x - 2\right)}$

So now we can apply the decomposition into partial fractions
$\frac{1}{\left({x}^{2}\right) \left(x - 2\right)} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{x - 2}$
$= \frac{A \left(x - 2\right) + B x \left(x - 2\right) + C {x}^{2}}{\left({x}^{2}\right) \left(x - 2\right)}$
Solving for A,B and C

$1 = A \left(x - 2\right) + B x \left(x - 2\right) + C {x}^{2}$
let $x = 2$$\implies$$1 = 4 C$$\implies$$C = \frac{1}{4}$
let $x = 0$$\implies$$1 = - 2 A$$\implies$$A = - \frac{1}{2}$
Coefficients of ${x}^{2}$$\implies$$0 = B + C$$\implies$$B = - \frac{1}{4}$
so we have, 1/((x^2)(x-2))=-1/(2x^2)-1/(4x)+1/(4(x-2)
So $1 - \frac{4}{\left({x}^{2}\right) \left(x - 2\right)} = 1 - \frac{2}{{x}^{2}} - \frac{1}{x} + \frac{1}{x - 2}$

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

$= x - \frac{2}{x} + \ln x - \ln \left(x - 2\right) + C$