Question

How do you integrate `(3x^3-5x^2-11x+9)/(x^2-2x-3)` using partial fractions?

Answer 1

`int(3x^3-5x^2-11x+9)/(x^2-2x-3)dx`

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

Explanation:

For converting to partial fractions, we must have degreeof numerator less than that of denominator. Hence as

`3x^3-5x^2-11x+9=3x(x^2-2x-3)+1(x^2-2x-3)+12` and hence

`(3x^3-5x^2-11x+9)/(x^2-2x-3)=3x+1+12/(x^2-2x-3)`

As `(x^2-2x-3)=(x-3)(x+1)`, let

`12/(x^2-2x-3)hArrA/(x-3)+B/(x+1)=(A(x+1)+B(x-3))/(x^2-2x-3)`

if we put `x=3`, `4A=12` i.e. `A=3`

and if we put `x=-1`, `-4B=12` or `B=-3`

Hence `(3x^3-5x^2-11x+9)/(x^2-2x-3)=3x+1+12/(x^2-2x-3)`

and `int(3x^3-5x^2-11x+9)/(x^2-2x-3)dx`

= `int(3x+1+3/(x-3)-3/(x+1))dx`

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