Question

How do you integrate `f(x)=(x^2-2x)/((x^2-3)(x-3)(x-8))` using partial fractions?

Answer 1

`int (x^2-2x)/[(x^2-3)(x-3)(x-8)*dx]`

=`5/122Ln(x-8)`+`(21+5sqrt3)/732ln(x+sqrt3)`-`1/10Ln(x-3)`-`(21-5sqrt3)/6ln(x-sqrt3)+C`

Explanation:

I decomposed into basic fractions,

`(x^2-2x)/[(x^2-3)(x-3)(x-8)]`

=`(Ax+B)/(x^2-3)+C/(x-3)+D/(x-8)`

After expanding denominator,

`(Ax+B)(x-3)(x-8)+C(x^2-3)(x-8)+D(x^2-3)(x-3)=x^2-2x`

Set `x=3`, `-30C=3`, so `C=-1/10`

Set `x=8`, `305D=48`, so `D=48/305`

Set `x=0`, `24B+24C+9D=0`, so `B=-C-3/8*D=5/122`

Set `x=2`, `14A+14B+14C+4D=-1`, so `A=-7/122`

Hence,

`int (x^2-2x)/[(x^2-3)(x-3)(x-8)*dx]`

=`-1/122int ((7x-5)*dx)/(x^2-3)-1/10int (dx)/(x-3)+5/122int (dx)/(x-8)`

=`5/122Ln(x-8)-1/10Ln(x-3)+C-1/122*int ((7x-5)*dx)/(x^2-3)`

Now I solved last integral,

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

=`int ((7x-5)*dx)/[(x+sqrt3)(x-sqrt3)]`

=`(21-5sqrt3)/6int (dx)/(x-sqrt3)-(21+5sqrt3)/6int (dx)/(x+sqrt3)`

=`(21-5sqrt3)/6ln(x-sqrt3)-(21+5sqrt3)/6ln(x+sqrt3)`

Thus,

`int (x^2-2x)/[(x^2-3)(x-3)(x-8)*dx]`

=`5/122Ln(x-8)+(21+5sqrt3)/732ln(x+sqrt3)-1/10Ln(x-3)-(21-5sqrt3)/6ln(x-sqrt3)+C`