# How do you integrate 2/[x^3-x^2 ] using partial fractions?

Jun 10, 2016

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

$= - 2 \ln | x | + \frac{2}{x} + 2 \ln | x - 1 | + C$.

#### Explanation:

Applying the method of partial fractions to rewrite the integral, we may write (denominator factors as linear factors, one repeated) :

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

=(Ax(x-1)+B(x-1)+Cx^2)/(x^2(x-1)

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

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

Hence : $A + C = 0 \implies A = - C$

$A - B = 0 \implies A = B$

$B = - 2$

$\therefore A = - 2 \mathmr{and} C = 2$

Hence we may rewrite the integral as :

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

$= - 2 \ln | x | + \frac{2}{x} + 2 \ln | x - 1 | + C$.