# How do you write the partial fraction decomposition of the rational expression (x^2 + 5x - 7 )/( x^2 (x+ 1)^2)?

Oct 23, 2016

The decomposition is

$\frac{{x}^{2} + 5 x - 7}{{x}^{2} {\left(x + 1\right)}^{2}} = - \frac{7}{x} ^ 2 + \frac{19}{x} - \frac{19}{x + 1} - \frac{11}{x + 1} ^ 2$

#### Explanation:

Let A,B,C,D be constants
The fraction decomposition of the rational expression

$\frac{{x}^{2} + 5 x - 7}{{x}^{2} {\left(x + 1\right)}^{2}} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{x + 1} + \frac{D}{x + 1} ^ 2$

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

So ${x}^{2} + 5 x - 7 = A {\left(x + 1\right)}^{2} + B x {\left(x + 1\right)}^{2} + C {x}^{2} \left(x + 1\right) + D {x}^{2}$

let $x = 0$, then $- 7 = A$

let $x = - 1$ then $- 11 = D$

Comparing coefficients of $x$

$5 = 2 A + B$ so $B = 5 - 2 A = 5 + 14 = 19$
Comparing the coefficients of ${x}^{2}$
$1 = A + 2 B + C + D$ so $C = 1 - A - 2 B - D = 1 + 7 - 38 + 11 = - 19$

so the final result is
$\frac{{x}^{2} + 5 x - 7}{{x}^{2} {\left(x + 1\right)}^{2}} = - \frac{7}{x} ^ 2 + \frac{19}{x} - \frac{19}{x + 1} - \frac{11}{x + 1} ^ 2$