Question

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

Answer 1

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

Explanation:

Let us find the partial fraction decomposition of the integrand:

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

This is an identity and valid `AA x in RR`
Put `x=-2 =>-6=-3A + 0 => A=2`
Put `x=1 => 3=0+B(3) => B=1`

So the partial fraction decomposition is:

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

And so the integral can be written as:

`int (3x)/((x+2)(x-1)) dx = int 2/(x+2) + 1/(x-1) dx`
`:. int (3x)/((x+2)(x-1)) dx = 2int 1/(x+2) dx+ int1/(x-1) dx`
`:. int (3x)/((x+2)(x-1)) dx = 2ln|x+2| + ln|x-1| + lnA` (`lnA`=constant)
`:. int (3x)/((x+2)(x-1)) dx = ln(A(x+2)^2|x-1|)`