# Question #85073

May 9, 2016

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

#### Explanation:

Performing the decomposition, we first set up the equation with the unknown constants.

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

Multiplying both sides by $x {\left(x - 1\right)}^{3}$, we get

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

$= \left(A + B\right) {x}^{3} + \left(- 3 A - 2 B + C\right) {x}^{2} + \left(3 A + B - C + D\right) x + \left(- A\right)$

Equating the coefficients of corresponding powers of $x$, we end up with the following system of equations:

$\left\{\begin{matrix}A + B = 0 \\ - 3 A - 2 B + C = 1 \\ 3 A + B - C + D = 0 \\ - A = 1\end{matrix}\right.$

From the fourth equation, we have $A = - 1$

Substituting that into the first equation and solving for $B$ gives $B = 1$

Substituting both of those into the second equation and solving for $C$ gives $C = 0$

Substituting all of those into the third equation and solving for $D$ gives $D = 2$

Thus, from our original equation, we have the decomposition

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