Question

How do you integrate `int (x-3)/(x^2-2x-5) dx` using partial fractions?

Answer 1

`int \ (x-3)/(x^2-2x-5) \ dx = 1/2ln |x^2-2x-5| -1/sqrt(6)ln | (x-1-sqrt(6))/(x-1+sqrt(6)) | + C`

Explanation:

We seek:

`I = int \ (x-3)/(x^2-2x-5) \ dx`

If we look at the quadratic on the numerator, we finds that if

`x^2-2x-5 = 0 => x = 1 +- sqrt(6)`

Therefore we find that we do not get "perfect" factors but rather:

`(x^2-2x-5 ) = (x - 1 - sqrt(6))(x - 1 + sqrt(6))`

So, although we could decompose into partial fractions it actually complicates the problem with the risk of an algebraic error.

A another approach, is to complete the square of the denominator:

`I = int \ (x-3)/((x-1)^2-1-5) \ dx`
`\ \ = int \ (x-3)/((x-1)^2-6) \ dx`

Substitute `u=x-1 => (du)/dx = 1`; then

`I = int \ (u-2)/(u^2-6) \ du`
`\ \ = int \ (u)/(u^2-6) - 2/(u^2-6) \ du`
`\ \ = int \ 1/2(2u)/(u^2-6) - 2/(u^2-sqrt(6)^2) \ du`

Both of these integrals are standard, so we can now integrate, giving:

`I = 1/2ln |u^2-6| -2 1/(2sqrt(6))ln | (u-sqrt(6))/(u+sqrt(6)) | + C`
`\ \ = 1/2ln |u^2-6| -1/sqrt(6)ln | (u-sqrt(6))/(u+sqrt(6)) | + C`

And, restoring the earlier substitution:

`I = 1/2ln |(x-1)^2-6| -1/sqrt(6)ln | ((x-1)-sqrt(6))/((x-1)+sqrt(6)) | + C`
`\ \ = 1/2ln |x^2-2x-5| -1/sqrt(6)ln | (x-1-sqrt(6))/(x-1+sqrt(6)) | + C`