Question

How do you integrate `int (1-2x^2)/((x-2)(x+7)(x+1))` using partial fractions?

Answer 1

`-1/9ln|(x-2)|+5/18ln|(x+7)|-1/6ln|(x+1)|+C`

Explanation:

`int(1-2x)/((x-2)(x+7)(x+1))dx`

consider the integrand. The denominators are all linear so we have a straightforward identity

`(1-2x)/((x-2)(x+7)(x+1))-+A/(x-2)+B/(x+7)+C/(x+1)`

multiply both sides by`((x-2)(x+7)(x+1))`

after cancelling we end up with

`(1-2x)-=A(x+7)(x+1)+B(x-2)(x+1)+C(x-2)(x+7)`

`"let " x=2`

`1-4=Axx9xx3+0+0`

`-3=27A`

`=>A=-1/9`

`"let "x=-7`

`1-(2xx-7)=0+Bxx-9xx-6+0`

`15=54B`

`=>B=15/54=5/18`

`"let "x=-1`

`1-(2xx-1)=0+0+Cxx(-1-2)(-1+7)`

`=>3=-18C`

`=>C=-1/6`

`:.int(1-2x)/((x-2)(x+7)(x+1))dx`

`int(-1/(9(x-2))+5/(18(x+7))-1/(6(x+1)))dx`

using`color(blue)(int(f'(x))/(f(x))dx=ln|f(x)|+C)`

we have

`-1/9ln|(x-2)|+5/18ln|(x+7)|-1/6ln|(x+1)|+C`