Question

How do you integrate `int 1/(x^2-x-20) dx` using partial fractions?

Answer 1

`1/9ln|x-5|-1/9ln|x+4|+C`

Explanation:

Factor the denominator of the integrand:

`x^2-x-20=(x-5)(x+4)`

Use partial fraction decomposition on the simplified integrand:

`1/((x-5)(x+4))=A/(x-5)+B/(x+4)`

Add up the right side:

`1/((x-5)(x+4))=(A(x+4)+B(x-5))/((x-5)(x+4))`

Equate numerators:

`1=A(x+4)+B(x-5)`

We need to find `A,B.` We can do this by plugging in values of `x` that send one term to `0` and keep the other:

`x=-4:`

`1=-9B, B=-1/9`

`x=5:`

`1=9A, A=1/9`

Thus, our integral becomes

`int(1/9)/(x-5)-(1/9)/(x+4)dx=1/9ln|x-5|-1/9ln|x+4|+C`

Answer 2

`1/9 * lnabs(x-5)- 1/9 * lnabs(x+4)`

Explanation:

`1/((x-5)(x+4)) = 1/(9*(x-5)) - 1/(9*(x+4))`

`int1/((x-5)(x+4)) = int1/(9*(x-5)) - 1/(9*(x+4))dx`

`int1/((x-5)(x+4)) = int1/(9*(x-5)) - 1/(9*(x+4))dx`

`int1/((x-5)(x+4)) = 1/9 * lnabs(x-5)- 1/9 * lnabs(x+4)`