Question

How do you integrate `int (3x)/((x + 2)(x - 1))` using partial fractions?

Answer 1

`int(3x)/((x+2)(x-1)) = ln(A(x+2)^2|x-1|)`

Explanation:

The partial fraction decomposition of the integrand will be of the form:

`(3x)/((x+2)(x-1)) = A/(x+2) + B/(x-1)`
`" " = (A(x-1) + B(x+2))/((x+2)(x-1))`
`:. 3x" " = A(x-1) + B(x+2)`

Put `x=-2 => -6=A(-3) => A=2`
Put `x=1 " "=> 3=B(3) " "=> B=1`

Hence,

`(3x)/((x+2)(x-1)) = 2/(x+2) + 1/(x-1)`

And so:

`int(3x)/((x+2)(x-1)) = int(2/(x+2) + 1/(x-1))dx`
`" "= 2int 1/(x+2)dx + int 1/(x-1)dx`
`" "= 2ln|x+2|+ln|x-1| + c`
`" "= ln(|x+2|)^2+ln|x-1| + lnA` (`c=lnA`)
`" "= ln(A(x+2)^2|x-1|)`