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

Jan 31, 2017

The answer is $= \frac{- \frac{1}{2}}{x} ^ 2 + \frac{- \frac{3}{4}}{x} + \frac{\frac{3}{4}}{x - 2}$

#### Explanation:

Let's perform the decomposition into partial fractions

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

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

As the denominators are the same, we can equalize the nimerators

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

Let $x = 0$, $\implies$, $1 = - 2 A$, $\implies$, $A = - \frac{1}{2}$

Let $x = 2$, $\implies$, $3 = 4 C$, $\implies$, $C = \frac{3}{4}$

Coefficients of ${x}^{2}$, $\implies$, $0 = B + C$

$C = - B = - \frac{3}{4}$

Therefore,

$\frac{x + 1}{{x}^{2} \left(x - 2\right)} = \frac{- \frac{1}{2}}{x} ^ 2 + \frac{- \frac{3}{4}}{x} + \frac{\frac{3}{4}}{x - 2}$