Question

How do you integrate `int (x-3x^2)/((x-6)(x-2)(x-5))` using partial fractions?

Answer 1

`int(x-3x^2)/((x-6)(x-2)(x-5))dx`

= `-121/8ln(x-6)+5/8ln(x-2)+35/4ln(x-5)+c`

Explanation:

Let us first convert `(x-3x^2)/((x-6)(x-2)(x-5))` into partial fractions.

`(x-3x^2)/((x-6)(x-2)(x-5))hArrA/(x-6)+B/(x-2)+C/(x-5)`

`(x-3x^2)/((x-6)(x-2)(x-5))hArr(A(x-2)(x-5)+B(x-6)(x-5)+C(x-6)(x-2))/((x-6)(x-2)(x-5))`

= `(A(x^2-7x+10)+B(x^2-11x+30)+C(x^2-8x+12))/((x-6)(x-2)(x-5))`

= `(x^2(A+B+C)-x(7A+11B+8C)+(10A+30B+12C))/((x-6)(x-2)(x-5))`

Hence `A+B+C=-3`, `7A+11B+8C=-1` and `10A+30B+12C=0`

Subtracting `7` times of first equation from second and `10` times first from third equation, we get

`4B+C=20` and `20B+C=30` which gives us `B=5/8` and `C=140/8=35/4` and putting these in `A+B+C=-3`, we get `A=-121/8`

Hence, `(x-3x^2)/((x-6)(x-2)(x-5))=-121/(8(x-6))+5/(8(x-2))+35/(4(x-5))`

and `int(x-3x^2)/((x-6)(x-2)(x-5))dx=int[-121/(8(x-6))+5/(8(x-2))+35/(4(x-5))]dx`

= `-121/8ln(x-6)+5/8ln(x-2)+35/4ln(x-5)+c`