Question

How do you find `int (5x+11)/(x^2+2x-35) dx` using partial fractions?

Answer 1

Do a partial fraction decomposition on `(5x+11)/(x^2+2x-35)` and simplify to get `2lnabs(x+7)+3lnabs(x-5)+C`.

Explanation:

Begin by factoring the denominator to simplify the integral to:
`int(5x+11)/((x+7)(x-5))dx`

Because the denominator contains only linear factors, our partial fraction decomposition will be of the form:
`A/(x+7)+B/(x-5)`

We can now set up the decomposition:
`(5x+11)/((x+7)(x-5))=A/(x+7)+B/(x-5)`
`(5x+11)/((x+7)(x-5))=(A(x-5)+B(x+7))/((x+7)(x-5))`

Equating the numerators and simplifying:
`5x+11=A(x-5)+B(x+7)`
Set `x=5` to find the value of `B`:
`5(5)+11=A(5-5)+B(5+7)->36=12B->B=3`
Similarly, set `x=-7` to find `A`:
`5(-7)+11=A(-7-5)+B(-7+7)->-24=-12A->A=2`

Our decomposition is therefore:
`(5x+11)/((x+7)(x-5))=2/(x+7)+3/(x-5)`

Putting this back into the integral and evaluating:
`int(5x+11)/((x+7)(x-5))dx=int2/(x+7)+3/(x-5)dx`
`color(white)(XX)=int2/(x+7)dx+int3/(x-5)dx`
`color(white)(XX)=2int1/(x+7)dx+3int1/(x-5)dx`
`color(white)(XX)=2lnabs(x+7)+3lnabs(x-5)+C`