# How do you express 1/((x²)(x² + 3x + 2))  in partial fractions?

Nov 15, 2016

$\frac{1}{{x}^{2} \left({x}^{2} + 3 x + 2\right)} = - \frac{3}{4 x} + \frac{1}{2 {x}^{2}} + \frac{1}{x + 1} - \frac{1}{4 \left(x + 2\right)}$

#### Explanation:

As $\frac{1}{{x}^{2} \left({x}^{2} + 3 x + 2\right)} = \frac{1}{{x}^{2} \left(x + 2\right) \left(x + 1\right)}$, let the partial fractions be given by

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

or $\frac{1}{{x}^{2} \left(x + 2\right) \left(x + 1\right)} \Leftrightarrow \frac{A x \left(x + 1\right) \left(x + 2\right) + B \left(x + 1\right) \left(x + 2\right) + C {x}^{2} \left(x + 2\right) + D {x}^{2} \left(x + 1\right)}{{x}^{2} \left(x + 2\right) \left(x + 1\right)}$

Now, if $x = - 1$, we have $C = 1$ and if $x = - 2$, we have $- 4 D = 1$ or $D = - \frac{1}{4}$ and if $x = 0$, we have $2 B = 1$ or $B = \frac{1}{2}$.

Now comparing the terms of ${x}^{3}$, we get $A + C + D = 0$ and hence
$A = - C - D = - 1 - \left(- \frac{1}{4}\right) = - \frac{3}{4}$

Hence $\frac{1}{{x}^{2} \left({x}^{2} + 3 x + 2\right)} = - \frac{3}{4 x} + \frac{1}{2 {x}^{2}} + \frac{1}{x + 1} - \frac{1}{4 \left(x + 2\right)}$

Nov 15, 2016

The answer is $= \frac{\frac{1}{2}}{x} ^ 2 + \frac{- \frac{3}{4}}{x} + \frac{- \frac{1}{4}}{x + 2} + \frac{1}{x + 1}$

#### Explanation:

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

The decomposition in partial fractions is

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

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

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

Let x=0, $\implies$$1 = 2 A$ ; $A = \frac{1}{2}$

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

Let $x = - 2$, $\implies$ $1 = - 4 C$ , $C = - \frac{1}{4}$

Coefficient of ${x}^{3}$, $\implies$, $0 = B + C + D$

$B = - C - D = \frac{1}{4} - 1 = - \frac{3}{4}$

$\frac{1}{{x}^{2} \left(x + 2\right) \left(x + 1\right)} = \frac{\frac{1}{2}}{x} ^ 2 + \frac{- \frac{3}{4}}{x} + \frac{- \frac{1}{4}}{x + 2} + \frac{1}{x + 1}$