# How do you express as a partial fraction (2x) / ((x-2)(x²+1)(x+1)²)?

Aug 23, 2015

$\frac{2 x}{\left(x - 2\right) \left({x}^{2} + 1\right) {\left(x + 1\right)}^{2}} = \frac{4}{45 \left(x - 2\right)} + \frac{1}{9 \left(x + 1\right)} + \frac{1}{3 {\left(x + 1\right)}^{2}} - \frac{x + 2}{5 \left({x}^{2} + 1\right)}$

#### Explanation:

The denominator is already factored, so we need:

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

Clear fractions to get:

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

$= A \left({x}^{4} + 2 {x}^{3} + 2 {x}^{2} + 2 x + 1\right) + B \left({x}^{4} - {x}^{3} - {x}^{2} - 2\right) + C \left({x}^{3} - 2 {x}^{2} + x - 2\right) + \left(D x + E\right) \left({x}^{3} - 3 x - 2\right)$

Do the algebra to regroup and equate coefficients. You'll get a system of 5 equations in 5 unknowns. Solve to find:

$A = \frac{4}{45}$, $\text{ } B = \frac{1}{9}$, $\text{ } C = \frac{1}{3}$, $\text{ } D = - \frac{1}{5}$, and $\text{ } E = - \frac{2}{5}$