Question

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

Answer 1

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

Explanation:

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`