# How do you write the partial fraction decomposition of the rational expression  (x^2+x)/((x+2)(x-1)^2)?

Nov 6, 2016

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

#### Explanation:

As the degree of numerator is less than that of denominator, we can proceed to write partial fractions as

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

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

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

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

Putting $x = 1$ and $x = - 2$ in (1), we get $3 C = 2$ and $9 A = 2$ i.e. $C = \frac{2}{3}$ and $A = \frac{2}{9}$.

Comparing terms of ${x}^{2}$ in (2), $A + B = 1$ i.e. $B = \frac{7}{9}$

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