Question

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

Answer 1

`-1/4 ln(x+1) -ln (x-2) +5/4 ln (x-3) +c`

Explanation:

Before integration, partial fractions can be done as explained below

The partial fractions would thus be

`-1/(4(x+1)) -1/(x-2)+5/(4(x-3)`

Integration is now simple `-1/4 int dx/(x+1) -int dx/(x-2) +5/4 int dx/(x-3)`

=`-1/4 ln(x+1) -ln (x-2) +5/4 ln (x-3) +c`