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

Feb 14, 2016

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

#### Explanation:

The decomposition that should be set up from this problem is

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

Note that since ${x}^{2}$ was in the denominator, both $x$ and ${x}^{2}$ are included in the denominators of the decomposition.

From here, we can get a common denominator of ${x}^{2} \left(x - 1\right)$ in each term.

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

The denominators are equal, so they can be removed, giving us the equation

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

Set $x = 1$ so both the $A$ and $B$ terms will equal $0$.

$1 + 1 = A \left(1\right) \left(0\right) + B \left(0\right) + C \left(1\right)$

ul(C=2

Set $x = 0$ so both the $A$ and $C$ terms will equal $0$.

$0 + 1 = A \left(0\right) \left(- 1\right) + B \left(- 1\right) + C \left(0\right)$

ul(B=-1

Now that we know the values of $B$ and $C$, we can arbitrarily set $x = 2$ and substitute in the known values of $B$ and $C$ to solve for $A$.

$2 + 1 = A \left(2\right) \left(1\right) + \left(- 1\right) \left(1\right) + 2 \left(4\right)$

$3 = 2 A + 7$

ul(A=-2

This leaves us with the decomposition of

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