Question

How do you integrate `(5x^2+7x-4)/(x^3+4x^2)` using partial fractions?

Answer 1

The answer is `=1/x+2lnx+3ln(x+4)+C`

Explanation:

let's simplify the denominator `x^3+4x^2=x^2(x+4)`
So `(x^2+7x-4)/(x^2(x+4))=A/x^2+B/x+C/(x+4)`

`:.5x^2+7x-4=A(x+4)+Bx(x+4)+Cx^2`
`-4=4A` `=>` `A=-1`
coefficients of x, `7=A+4B` `=>` `B=2`
Coefficients of x^2, `5=B+C` `=>` `C=3`
`:. int((x^2+7x-4)dx)/(x^2(x+4))=int-dx/x^2+int(2dx)/x+int(3dx)/(x+4)`
`=1/x+2lnx+3ln(x+4)+C`