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

May 26, 2017

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

#### Explanation:

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Coeficients of ${x}^{3}$

$1 = A + C$

Let $x = 0$, $\implies$, $1 = 2 B + D$

Coefficients of ${x}^{2}$

$1 = B + D$

Coefficients of $x$

$2 = 2 A + C$

Solving for $A$, $B$, $C$ and $D$ from the 4 equations

$2 = 2 A + 1 - A$, $\implies$, $A = 1$

$C = 1 - A = 1 - 1 = 0$

$1 = 2 B + 1 - B$, $\implies$, $B = 0$

$1 = B + D$, $\implies$, $D = 1$

Therefore,

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