# How do you use partial fraction decomposition to decompose the fraction to integrate (x+2)/(x^3-2x^2+x)?

Aug 3, 2015

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

#### Explanation:

Factor the denominator:

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

Find $A , B , \text{ and } C$ to solve:

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

That is:

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

(When I was learning this is was clearest to me to get a common denominator on the left. So I'll show it that way.)

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

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

Regrouping and setting the numerators equal to each other, we get:

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

$= 0 {x}^{2} + 1 x + 2$

So

$A + B = 0$
$- 2 A - B + C = 1$
$A = 2$

From the first and last equation we can see that: $A = 2$ and $B = - 2$.

Substituting in the middle equation, we find that $C = 3$

So:

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