How do you integrate `int(x+1)/((x-5)(x+8)(x+4))` using partial fractions?

Luckily, in your case, the denominator is factorized already.

So, you don't need to prepare anything and can start doing the partial fraction decomposition immediately.

Your goal is to find `A`, `B`, `C` so that

`(x+1)/ ((x-5)(x+8)(x+4)) = A / (x-5) + B / (x+8) + C / (x+4)`

To do so, let's multiply the equation with the denominator first:

`x + 1 = A(x+8)(x+4) + B (x-5)(x+4) + C (x-5)(x+8)`

... expand the products...

`x + 1 = A * x^2 + A * 12x + A * 32 + B * x^2 - B * x - B * 20 + C * x^2 + C * 3x - C * 40`

... "sort" by `color(green)(x^2)`, `color(red)(x)` and `color(blue)("numbers")`...

`color(red)(x) + color(blue)(1) = color(green)(Ax^2) + color(red)(12Ax) + color(blue)(32A) + color(green)(Bx^2) color(red)(- B x) color(blue)(- 20B) +color(green)(C x^2) + color(red)(3Cx) color(blue)(- 40C)`

Now, in order to solve this equation for `A`, `B` and `C`, all `x^2` terms must match, all `x` terms must match and all terms without `x` must match. Thus, we can formulate three equations:

`{ (0 = A + B + C color(white)(xxxxxxxxxx) color(green)(x^2) " terms"), (1 = 12A - B + 3C color(white)(xxxxxxxx) color(red)(x) " terms"), (1 = 32A - 20B- 40 C color(white)(xxxxx) color(blue)("without " x)):}`

The solution of this linear equation system is

`A = 2/39`, `" "B = -7/52`, `" "C = 1/12`

Thus, your integral can be transformed into:

`int (x+1)/ ((x-5)(x+8)(x+4)) "d"x`

`= int (2/39 * 1/(x-5) - 7/52 * 1 / (x+8) + 1/12 * 1/(x+4)) "d"x`

`= 2/39 int 1/(x-5) "d"x - 7/52 int 1 / (x+8) "d" x + 1/12 int 1/(x+4) "d"x`

`= 2/39 ln abs(x-5) - 7/52 ln abs (x+8) + 1/12 ln abs(x+4) + c`