How do you integrate `int frac{2x-4}{(x-4)(x+3)(x-6)} dx` using partial fractions?

Applying partial fraction decomposition:

`(2x-4)/((x-4)(x+3)(x-6)) = A/(x-4)+B/(x+3)+C/(x-6)`

`=> 2x-4 = A(x+3)(x-6) + B(x-4)(x-6) + C(x-4)(x+3)`

`=> 2x-4 = (A+B+C)x^2 + (-3A -10B -C)x +(-18A +24B -12C)`

Equating equivalent coefficients gives us the system

`{(A+B+C = 0), (-3A -10B - C = 2), (-18A +24B -12C = 4):}`

Solving, we get

`{(A = -2/7), (B=-10/63), (C = 4/9):}`

Thus, substituting back in,

`(2x-4)/((x-4)(x+3)(x-6)) = -(2/7)/(x-4)-(10/63)/(x+3)+(4/9)/(x-6)`

and so, integrating,

`int(2x-4)/((x-4)(x+3)(x-6))dx`

`= int(-(2/7)/(x-4)-(10/63)/(x+3)+(4/9)/(x-6))dx`

`= -(2/7)int1/(x-4)dx -(10/63)int1/(x+3)dx`
`+(4/9)int1/(x-6)dx`

`= -2/7ln|x-4| - 10/63ln|x+3| + 4/9ln|x-6|+C`

First, you need to write out the partial fractions. The denominator has already been factorized for you.

`frac{2x - 4}{(x - 4)(x + 3)(x - 6)} -= frac{A}{x - 4} + frac{B}{x + 3} + frac{C}{x - 6}`

Where `A`, `B` and `C` are constants to be determined. Note that the sign `-=` means that the equality holds true for all possible values of `x`. Get rid of the denominators by multiplying both sides with `(x - 4)(x + 3)(x - 6)`.

`2x - 4 -= A(x + 3)(x - 6) + B(x - 4)(x - 6) + C(x - 4)(x + 3)`

When `x = 4`

`2(4) - 4 = A(4 + 3)(4 - 6)`

`A = -2/7`

When `x = -3`

`2(-3) - 4 = B(-3 - 4)(-3 - 6)`

`B = -10/63`

When `x = 6`

`2(6) - 4 = C(6 - 4)(6 + 3)`

`C = 4/9`

Therefore,

`frac{2x - 4}{(x - 4)(x + 3)(x - 6)} -= -frac{2/7}{x - 4} - frac{10/63}{x + 3} + frac{4/9}{x - 6}`.

Now, we proceed with the integration.

`int frac{2x - 4}{(x - 4)(x + 3)(x - 6)} dx = int (-frac{2/7}{x - 4} - frac{10/63}{x + 3} + frac{4/9}{x - 6}) dx`

`= -2/7 int frac{1}{x - 4} dx - 10/63 int frac{1}{x + 3} dx + 4/9 int frac{1}{x - 6} dx`

`= -2/7 ln|x - 4| - 10/63 ln|x + 3| + 4/9 ln|x - 6| + C`,

where `C` is the constant of integration.