# How do you express (6x^2+1)/(x^2(x-1)^2) in partial fractions?

Nov 25, 2017

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

#### Explanation:

$\frac{6 {x}^{2} + 1}{{x}^{2} {\left(x - 1\right)}^{2}} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x - 1} + \frac{D}{x - 1} ^ 2$

So:

$6 {x}^{2} + 1 = A x {\left(x - 1\right)}^{2} + B {\left(x - 1\right)}^{2} + C {x}^{2} \left(x - 1\right) + D {x}^{2}$

Putting $x = 1$ we get:

$7 = D$

Putting $x = 0$ we get:

$1 = B$

So:

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

$\textcolor{w h i t e}{6 {x}^{2} + 1} = A x \left({x}^{2} - 2 x + 1\right) + \left({x}^{2} - 2 x + 1\right) + C {x}^{2} \left(x - 1\right) + 7 {x}^{2}$

$\textcolor{w h i t e}{6 {x}^{2} + 1} = A \left({x}^{3} - 2 {x}^{2} + x\right) + \left({x}^{2} - 2 x + 1\right) + C \left({x}^{3} - {x}^{2}\right) + 7 {x}^{2}$

$\textcolor{w h i t e}{6 {x}^{2} + 1} = \left(A + C\right) {x}^{3} + \left(8 - 2 A - C\right) {x}^{2} + \left(A - 2\right) x + 1$

Hence:

$A = 2 \text{ }$ and $\text{ } C = - 2$

So:

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