Question

How do you integrate `int (x-3x^2)/((x-7)(x-5)(x+4))` using partial fractions?

Answer 1

`int (x-3x^2)/((x-7)(x-5)(x+4)) dx`

`= -70/11 ln abs(x-7)+35/9 ln abs(x-5)-52/99 ln abs(x+4) + C`

Explanation:

`(x-3x^2)/((x-7)(x-5)(x+4)) = A/(x-7)+B/(x-5)+C/(x+4)`

Using Heaviside's cover up method, we find:

`A = ((7)-3(7)^2)/(((7)-5)((7)+4)) = (7-147)/(2*11) = -140/22 = -70/11`

`B = ((5)-3(5)^2)/(((5)-7)((5)+4)) = (5-75)/((-2)*9) = (-70)/(-18) = 35/9`

`C = ((-4)-3(-4)^2)/(((-4)-7)((-4)-5)) = (-4-48)/((-11)(-9)) = -52/99`

So:

`int (x-3x^2)/((x-7)(x-5)(x+4)) dx`

`= int -70/(11(x-7))+35/(9(x-5))-52/(99(x+4)) dx`

`= -70/11 ln abs(x-7)+35/9 ln abs(x-5)-52/99 ln abs(x+4) + C`

(Note: The `C` here stands for the integration constant, not the coefficient we calculated).