How do you write the partial fraction decomposition of the rational expression  (x^2)/(x+1)^3?

Mar 29, 2017

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

Explanation:

${x}^{2} / {\left(x + 1\right)}^{3} = \frac{A}{x + 1} + \frac{B}{x + 1} ^ 2 + \frac{C}{x + 1} ^ 3$

$\textcolor{w h i t e}{{x}^{2} / {\left(x + 1\right)}^{3}} = \frac{A {\left(x + 1\right)}^{2} + B \left(x + 1\right) + C}{x + 1} ^ 3$

$\textcolor{w h i t e}{{x}^{2} / {\left(x + 1\right)}^{3}} = \frac{A \left({x}^{2} + 2 x + 1\right) + B \left(x + 1\right) + C}{x + 1} ^ 3$

$\textcolor{w h i t e}{{x}^{2} / {\left(x + 1\right)}^{3}} = \frac{A {x}^{2} + \left(2 A + B\right) x + \left(A + B + C\right)}{x + 1} ^ 3$

Equating coefficients we have:

$\left\{\begin{matrix}A = 1 \\ 2 A + B = 0 \\ A + B + C = 0\end{matrix}\right.$

Hence:

$\left\{\begin{matrix}A = 1 \\ B = - 2 \\ C = 1\end{matrix}\right.$

So:

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