How do you express (x^3-x^2+1) / (x^4-x^3) in partial fractions?

May 22, 2017

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

Explanation:

The denominator is

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

Let's perform the decomposition into partial fractions

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

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

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

The denominators are the same, we compare the numerators

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

Let $x = 0$, $\implies$, $1 = - A$, $\implies$, $A = - 1$

Let $x = 1$, $\implies$, $1 = D$

Coefficients of ${x}^{2}$,

$- 1 = B - C$

Coefficients of ${x}^{3}$,

$1 = C + D$, $\implies$, $C = 1 - D = 1 - 1 = 0$

$B = C - 1 = 0 - 1 = - 1$

Therefore,

$\frac{{x}^{3} - {x}^{2} + 1}{{x}^{4} - {x}^{3}} = - \frac{1}{x} ^ 3 - \frac{1}{{x}^{2}} + \frac{0}{x} + \frac{1}{x - 1}$