Question

How do you express `(x^4+6)/(x^5+7x^3)` in partial fractions?

Answer 1

The result is `=(-6/49)/x+(6/7)/x^3+(55/49x)/(x^2+7)`

Explanation:

Let's factorise the denominator `x^5+7x^3=x^3(x^2+7)`
So `(x^4+6)/(x^3(x^2+7))=A/x+B/x^2+C/x^3+(Dx+E)/(x^2+7)`
`=(Ax^2(x^2+7)+Bx(x^2+7)+C(x^2+7)+(Dx+E)(x^2+7))/(x^3(x^2+7))`
`x^4+6=Ax^2(x^2+7)+Bx(x^2+7)+C(x^2+7)+(Dx+E)x^3`
`x=0` `=>` `6=7C` `=>` `C=6/7`
coefficients `x^4` `=>` `1=A+D`
coefficients `x^3` `=>` `0=B+E`
coefficients `x^2` `=>` `0=7A+C`
coefficients of `x` `=>` `0=7B`
`E=0`
`A=-C/7=-6/49`
`B=0`
`D=1-A=1+6/49=55/49`
Finally we have
`(x^4+6)/(x^3(x^2+7))=(-6/49)/x+(6/7)/x^3+(55/49x)/(x^2+7)`