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

Dec 7, 2016

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

#### Explanation:

Assume that there exist $A , B \mathmr{and} C$ such that:
${x}^{2} / {\left(x - 1\right)}^{3} \equiv \frac{A}{\left(x - 1\right)} + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1} ^ 3$.
To find $A$, $B$ and $C$ substitute any three different numbers other than $- 1$ and solve the resulting set of three linear equations. For example, set $x = 0$, $2$ and $- 1$:
$0 = \frac{A}{-} 1 + B + \frac{C}{-} 1$
$4 = A + B + C$
$- \frac{1}{8} = \frac{A}{-} 2 + \frac{B}{4} - \frac{C}{8}$
giving $A = 1$, $B = 2$ and $C = 1$.
So the required integral is:
$\int \frac{1}{x - 1} + \frac{2}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 3 \mathrm{dx}$
Alternatively, instead of using partial fractions, substitute $u = x - 1$ to get $\int \frac{1}{u} + \frac{2}{u} ^ 2 + \frac{1}{u} ^ 3 \mathrm{du}$.