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

Oct 29, 2016

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

#### Explanation:

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

So now we can solve for $A , B , \mathmr{and} C$
$x - 1 = A x \left(x + 1\right) + B \left(x + 1\right) + C {x}^{2}$
let x=-1, $- 2 = C$$\implies$$C = - 2$
Coefficients of ${x}^{2}$, $0 = A + C$$\implies$$A = 2$
Coefficients of $x$, $1 = A + B$$\implies$$B = - 1$

And finally, we have
$\frac{x - 1}{{x}^{3} + {x}^{2}} = \frac{2}{x} - \frac{1}{x} ^ 2 - \frac{2}{x + 1}$