# Find the partial fraction decomposition of (x^3+x^2+x+2)/(x^4+x^2)?

Jan 23, 2017

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

#### Explanation:

As factors of ${x}^{4} + {x}^{2} = {x}^{2} \left({x}^{2} + 1\right)$, partial fraction decomposition of $\frac{{x}^{3} + {x}^{2} + x + 2}{{x}^{4} + {x}^{2}}$ will be of of the form

$\frac{{x}^{3} + {x}^{2} + x + 2}{{x}^{4} + {x}^{2}} \Leftrightarrow \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C x + D}{{x}^{2} + 1}$

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

= $\frac{A {x}^{3} + A x + B {x}^{2} + B + C {x}^{3} + D {x}^{2}}{{x}^{4} + {x}^{2}}$

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

Comparing the coefficients, we get $B = 2$

and as $B + D = 1$ i.e. $D = - 1$. Further, $A = 1$,

as $A + C = 1$ we have $C = 0$

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