How do you find `int (x - 4 ) /((x-1)(x+2)(x-3)) dx` using partial fractions?

Since the denominator is already factored into distinct linear factors, we are looking for a partial fraction decompositionof the form:

`(x-4)/((x-1)(x+2)(x-3)) = A/(x-1)+B/(x+2)+C/(x-3)`

`=(A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2))/((x-1)(x+2)(x-3))`

`=(A(x^2-x-6)+B(x^2-4x+3)+C(x^2+x-2))/((x-1)(x+2)(x-3))`

`=((A+B+C)x^2+(-A-4B+C)x+(-6A+3B-2C))/((x-1)(x+2)(x-3))`

Equating coefficients, we get the system of simulataneous linear equations:

`{ (A+B+C = 0), (-A-4B+C=1), (-6A+3B-2C=-4) :}`

Add multiples of the first equation to the second and third equations to get:

`{ (A+B+C = 0), (-3B+2C=1), (9B+4C=-4) :}`

Add three times the second equation to the third equation to get:

`{ (A+B+C = 0), (-3B+2C=1), (10C=-1) :}`

Divide the third equation by `10` to get:

`{ (A+B+C = 0), (-3B+2C=1), (C=-1/10) :}`

Subtract multiples of the third equation from the other equations to get:

`{ (A+B = 1/10), (-3B=6/5), (C=-1/10) :}`

Divide the second equation by `-3` to get:

`{ (A+B = 1/10), (B=-2/5), (C=-1/10) :}`

Subtract the second equation from the first to get:

`{ (A = 1/2), (B=-2/5), (C=-1/10) :}`

So:

`(x-4)/((x-1)(x+2)(x-3)) = 1/(2(x-1))-2/(5(x+2))-1/(10(x-3))`

So:

`int (x-4)/((x-1)(x+2)(x-3)) dx`

`= int 1/(2(x-1))-2/(5(x+2))-1/(10(x-3)) dx`

`= 1/2 ln abs(x-1) - 2/5 ln abs(x+2) - 1/10 ln abs(x-3) + C`