# How do you use partial fraction decomposition to decompose the fraction to integrate (x^3+x^2+x+2)/(x^4+x^2)?

Sep 25, 2015

See the explanation.

#### Explanation:

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

$\frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C x + D}{{x}^{2} + 1} =$

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

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

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

$A + C = 1$
$B + D = 1$
$A = 1$
$B = 2$

$C = 1 - A = 0$
$D = 1 - B = - 1$

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

$= \ln | x | - \frac{2}{x} - \arctan x + C$