# How do you write the partial fraction decomposition of (2x^3-x^2+x+5)/(x(x+1)^2)?

Dec 22, 2016

The answer is $= 2 + \frac{5}{x} - \frac{1}{x + 1} ^ 2 - \frac{10}{x + 1}$

#### Explanation:

Since the degree of the numerator is $=$ to the degree of the denominator, we must perform a long division

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

Now, we perform the long division

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} - {x}^{2} + x + 5$$\textcolor{w h i t e}{a a a a}$∣${x}^{3} + 2 {x}^{2} + x$

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} + 4 {x}^{2} + 2 x$$\textcolor{w h i t e}{a a a a a}$∣$2$

$\textcolor{w h i t e}{a a a a a a}$$0 - 5 {x}^{2} - x + 5$

Therefore,

$\frac{2 {x}^{3} - {x}^{2} + x + 5}{x {\left(x + 1\right)}^{2}} = 2 + \frac{- 5 {x}^{2} - x + 5}{x {\left(x + 1\right)}^{2}}$

We can make the partial fraction decomposition

$\frac{- 5 {x}^{2} - x + 5}{x {\left(x + 1\right)}^{2}} = \frac{A}{x} + \frac{B}{x + 1} ^ 2 + \frac{C}{x + 1}$

=(A(x+1)^2+Bx+Cx(x+1))/((x(x+1)^2)

So,

$- 5 {x}^{2} - x + 5 = A {\left(x + 1\right)}^{2} + B x + C x \left(x + 1\right)$

Let $x = 0$, $\implies$, $5 = A$

Let $x = - 1$, $\implies$, $1 = - B$, $\implies$, $B = - 1$

Coefficients of ${x}^{2}$, $\implies$,$- 5 = A + C$, $\implies$, $C = - 10$

Finally we get

$\frac{2 {x}^{3} - {x}^{2} + x + 5}{x {\left(x + 1\right)}^{2}} = 2 + \frac{5}{x} - \frac{1}{x + 1} ^ 2 - \frac{10}{x + 1}$