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

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

#### Explanation:

From the given $\frac{x + 1}{{x}^{2} \left(x - 1\right)}$, we set up the equations with the needed unknown constants represented by letters

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

The Least Common Denominator is ${x}^{2} \left(x - 1\right)$.

Simplify

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

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

Expand the numerator of the right side of the equation

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

Rearrange from highest degree to the lowest degree the terms of the numerator at the right side

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

Now let us fix the numerators on both sides so that the numerical coefficients are matched.

(0*x^2+1*x+1*x^0)/(x^2(x-1))=((B+C)x^2+(A-B)x-A*x^0)/(x^2(x-1)

We can now have the equations to solve for A, B, C

$B + C = 0$
$A - B = 1$
$- A = 1$

Simultaneous solution results to
$A = - 1$
$B = - 2$
$C = 2$

We now have the final equivalent
$\frac{x + 1}{{x}^{2} \left(x - 1\right)} = - \frac{1}{x} ^ 2 - \frac{2}{x} + \frac{2}{x - 1}$

God bless....I hope the explanation is useful.