# How do you express x^2 / (x-1)^3 in partial fractions?

Jun 1, 2016

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

#### Explanation:

The expansion for ${x}^{2} / {\left(x - 1\right)}^{3}$ is given by

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

reducing the right member to common denominator we have

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

Equating the numerators we have

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

giving the following conditions

{ (A=1), (B-2A=0), (A-B+C=0) :}

solving for $A , B , C$ we get

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