Question

How do you integrate `int (x-5) / (x^2(x+1))` using partial fractions?

Answer 1

`6ln|x|+5/x-6ln|x+1|+C`

Explanation:

Decompose into partial fractions:
`(x-5)/(x^2(x+1))-=(Ax+B)/x^2-6/(x+1)`
giving
`x-5-=(Ax+B)(x+1)-6x^2`
Regrouping:
`0x^2+x^1-5x^0-=(A-6)x^2+(A+B)x^1+Bx^0`
Equating powers of `x`:
`A=6`, `B=-5`
giving the integrand as
`(6x-5)/x^2-6/(x+1)`
`=6/x-5/x^2-6/(x+1)`
The `6` in the partial fractions comes from the cover-up rule, which avoids getting three simultaneous equations (one for each power of `x` in the identity) to solve.