Question

How do you integrate `int (x+4)/(x^2+4x-5) dx` using partial fractions?

Answer 1

`1/6 * ln |x+5| + 5/6 * ln | x-1 |`

Explanation:

1) First step is to factorize the denominator of your fraction completely:

`x^2 + 4x - 5 = (x + 5)(x - 1)`

In case you don't know how to do this:
- solve the quadratic equation `x^2 + 4x - 5 = 0`
- find the solutions `x = -5` and `x = 1`
- your factorization is `(x - "solution"_1)(x - "solution"_2)`

2) Now that you have the factorization, you need to decompose your fraction into partial fractions.

This means that you are searching for `A` and `B` so that

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

... multiply both sides of the equation with `(x+5)(x-1)`...

`<=> x + 4 = A(x-1) + B(x+5)`
`<=> x + 4 = A * x - A + B * x + 5B`
`<=> color(blue)(x) + color(red)(4) = color(blue)(A * x) color(white)(x) color(red)(- A) + color(blue)(B * x) + color(red)(5B)`

To solve this equation, you need to compare the `color(blue) x` terms and the `color(red)"constant"` terms:

`{ (1 = A + B), (4 = - A + 5B) :}`

The solution of this linear equation system is `B = 5/6`, `A = 1/6`.

This means that you have succeded in the partial fraction decomposition:

`(x+4)/((x+5)(x-1)) = 1/6 * 1/(x+5) + 5/6 * 1/(x-1)`

3) The only thing left to do is solving the integral. :-)

`int (x+4)/(x^2 + 4x - 5) "d"x = int (x+4)/ ((x+5)(x-1)) "d"x`

`color(white)(xxxxxxxxxxx) = int 1/6 * 1/(x+5) + 5/6 * 1/(x-1) "d"x`

`color(white)(xxxxxxxxxxx) = 1/6 * int 1/(x+5) "d"x + 5/6 * int 1/(x-1) "d"x`

`color(white)(xxxxxxxxxxx) = 1/6 * ln |x+5| + 5/6 * ln | x-1 |`