# How do you express (x^3+x^2+x+2)/(x^4+x^2) in partial fractions?

Aug 14, 2017

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

#### Explanation:

 (x^3+x^2+x+2)/(x^4+x^2) = (x^3+x^2+x+2)/(x^2(x^2+1)

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

Multiplying both sides by ${x}^{2} \left({x}^{2} + 1\right)$ we get

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

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

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

Equating with powers of $x$ and constant term we get

$B = 2 , A = 1 , B + D = 1 , A + C = 1 \therefore C = 0 , D = - 1 \therefore$

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