# What is the partial-fraction decomposition of (x^2+2x+7)/(x(x-1)^2)?

Jun 24, 2015

$\frac{{x}^{2} + 2 x + 7}{x {\left(x - 1\right)}^{2}} = \frac{7}{x} - \frac{6}{x - 1} + \frac{10}{x - 1} ^ 2$

#### Explanation:

The factors on the denominator are obvious. They are all linear, but one of them is a double factor. So we want $A , B , \mathmr{and} C$ so that:

$\frac{{x}^{2} + 2 x + 7}{x {\left(x - 1\right)}^{2}} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x - 1} ^ 2$

Combining the rations on the right, we get a numerator pf:
$A \left({x}^{2} - 2 x + 1\right) + B \left({x}^{2} - x\right) + C x = A {x}^{2} - 2 A x + A + B {x}^{2} - B x + C x$

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

Setting the coefficients equal to those of the original numerator, ${x}^{2} + 2 x + 7$, we get:

$A + B = 1$
$- 2 A - B + C = 2$
$A = 7$

It is immediate that $A = 7$ and from that and the first equation (the coefficients of ${x}^{2}$), we get $B = - 6$. Substituting in the middle equation and solving for $C$, we get $C = 10$.

$\frac{{x}^{2} + 2 x + 7}{x {\left(x - 1\right)}^{2}} = \frac{7}{x} - \frac{6}{x - 1} + \frac{10}{x - 1} ^ 2$

It is a good idea to check the answer by getting the common denominator. (I did that on paper, but I'm not going to type it up.)