Question

How do you integrate `int (x-9)/((x+3)(x-7)(x-5))` using partial fractions?

Answer 1

`-3/20 ln abs (x+3) -1/10 ln abs (x-7) + 1/4 ln abs (x-5) + C`

Explanation:

As the other answer was unfortunately deleted, I will answer this question.

To compute a partial fractions decomposition means to find `A`, `B` and `C` such as

`(x-9)/((x+3)(x-7)(x-5)) = A / (x+3) + B / (x-7) + C / (x-5)`

... multiply both sides with the denominator `(x+3)(x-7)(x-5)`...

`<=>" " x - 9 = A(x-7)(x-5) + B(x+3)(x-5) + C (x+3)(x-7)`

`<=>" " x - 9 = A(x^2 - 12x + 35) + B(x^2 - 2x - 15) + C (x^2 - 4x - 21)`

`<=>" " x - 9 = A * x^2 - A * 12x + A * 35 + B * x^2 - B * 2x - B * 15 + C * x^2 - C * 4x - C * 21`

... identify the `color(red)(x^2)` terms, the `color(blue)(x)` terms and the `"constant"` terms:

`<=>" " color(red)(0 * x^2) + color(blue)(x) - 9 = color(red)(A * x^2) color(blue)(- 12A * x) + 35A + color(red)(B * x^2) color(blue)(- 2B * x) - 15B + color(red)(C * x^2) color(blue)(- 4C * x) - 21C`

This leads us to the following linear equation system:

`{ (" " (I) " "0 = " "A + B " "+C " " color(red)(x^2) " terms"), (" " (II) " "1 = -12A - 2 B " "- 4C " " color(blue)(x) " terms"), (" " (III) -9 = " "35A - 15B - 21C " constant terms") :}`

Now, let's solve this linear equation system.

To eliminate `B`, let's compute `(IV) = 2 * (I) + (II)` and `(V) = 15 * (I) + (III)`:

`(IV) " " 1 = -10 A - 2C`
`(V) " " -9 = " "50 A - 6C`

Now, we can compute `(VI) = 5 * (IV) + (V)` to eliminate `A`:

`(VI)" " -4 = -16 C`

Thus, we can solve `(VI)` for `C` and find `C = 1/4`.

Plugging `C` into `(IV)` gives us `A = -3/20` and finally, plugging `A` and `C` into `(I)` gives us `B = - 1/10`.

Now, we can perform the partial fraction decomposition and solve the integral:

`int (x-9)/((x+3)(x-7)(x-5)) "d"x`

`= int [-3/20 * 1 / (x+3) -1/10 * 1/ (x-7) + 1/4 * 1/ (x-5) ] "d"x`

`= -3/20 int [ 1 / (x+3)"d"x -1/10 int 1 / (x-7) "d"x + 1/4 int 1/ (x-5) "d"x`

`= -3/20 ln abs (x+3) -1/10 ln abs (x-7) + 1/4 ln abs (x-5) + C`