# How do you express (x^2 + 5x - 7 )/( x^2 (x+ 1)^2) in partial fractions?

Nov 4, 2016

The answer is $= - \frac{7}{x} ^ 2 + \frac{19}{x} - \frac{11}{x + 1} ^ 2 - \frac{19}{x + 1}$

#### Explanation:

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

${x}^{2} + 5 x - 7 = A {\left(x + 1\right)}^{2} + B x {\left(x + 1\right)}^{2} + C {x}^{2} + D {x}^{2} \left(x + 1\right)$
let $x = 0$ $\implies$$- 7 = A$
$x = - 1$$\implies$$- 11 = C$
coefficentsof ${x}^{2}$ $\implies$$1 = A + 2 B + C + D$
coefficients of $x$$\implies$$5 = 2 A + B$$\implies$$B = 19$
$\therefore D = - 19$
So, $\frac{{x}^{2} + 5 x - 7}{{x}^{2} {\left(x + 1\right)}^{2}} = - \frac{7}{x} ^ 2 + \frac{19}{x} - \frac{11}{x + 1} ^ 2 - \frac{19}{x + 1}$