Question

How do you integrate `int(x+1)/((2x-4)(x-4)(x-7))` using partial fractions?

Answer 1

`3/20Ln(Abs(x-2))-5/12(Abs(x-4))+4/15Ln(Abs(x-7))+C`

Explanation:

`(x+1)/[(2x-4)*(x-4)(x-7)]`

=`1/2*(x+1)/[(x-2)*(x-4)(x-7)]`

I decomposed `(x+1)/[(x-2)*(x-4)(x-7)]` into basic fractions,

`(x+1)/[(x-2)*(x-4)(x-7)]`

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

After expanding denominator,

`A(x-4)(x-7)+B(x-2)(x-7)+C(x-2)(x-4)=x+1`

Set `x=2`, `10A=3`, so `A=3/10`

Set `x=4`, `-6B=5`, so `B=-5/6`

Set `x=7`, `15C=8`, so `C=8/15`

Thus,

`int (x+1)/[(2x-4)*(x-4)(x-7)]*dx`

=`1/2int (x+1)/[(2x-4)*(x-4)(x-7)]*dx`

=`1/2*3/10int (dx)/(x-2)-1/2*5/6int (dx)/(x-4)+1/2*8/15int (dx)/(x-7)`

=`3/20int (dx)/(x-2)-5/12int (dx)/(x-4)+4/15int (dx)/(x-7)`

=`3/20Ln(Abs(x-2))-5/12(Abs(x-4))+4/15Ln(Abs(x-7))+C`