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

Jan 17, 2017

The answer is $= \frac{\frac{2}{9}}{x + 2} + \frac{\frac{7}{9}}{x - 1} + \frac{\frac{2}{3}}{x - 1} ^ 2$

#### Explanation:

Let's do the decomposition into partial fractions

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

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

We equalise the denominators

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

Let $x = - 2$, $\implies$,$2 = 9 A$, $\implies$, $A = \frac{2}{9}$

Let $x = 1$, $\implies$, $2 = 3 C$, $\implies$, $C = \frac{2}{3}$

Coefficients of ${x}^{2}$

$1 = A + B$, $\implies$, $B = 1 - A = 1 - \frac{2}{9} = \frac{7}{9}$

So,

$\frac{{x}^{2} + x}{\left(x + 2\right) {\left(x - 1\right)}^{2}} = \frac{\frac{2}{9}}{x + 2} + \frac{\frac{7}{9}}{x - 1} + \frac{\frac{2}{3}}{x - 1} ^ 2$