Question

How do you integrate `int (13x) / (6x^2 + 5x - 6)` using partial fractions?

Answer 1

`2/3ln(abs(3x-2))+3/2ln(abs(2x+3))+C`

Explanation:

Note that

`6x^2+5x-6=6x^2+9x-4x-6`

`=3x(2x+3)-2(2x+3)`

`=(3x-2)(2x+3)`

The partial fraction decomposition can then be set up as:

`(13x)/((3x-2)(2x+3))=A/(3x-2)+B/(2x+3)`

Multiplying both sides by `(3x-2)(2x+3)`, we see that

`13x=A(2x+3)+B(3x-2)`

Letting `x=-3/2`:

`13(-3/2)=A(2(-3/2)+3)+B(3(-3/2)-2)`

`-39/2=A(-3+3)+B(-9/2-2)`

`-39/2=B(-13/2)`

`3=B`

Letting `x=2/3`:

`13(2/3)=A(2(2/3)+3)+B(3(2/3)-2)`

`26/3=A(4/3+3)+B(2-2)`

`26/3=A(13/3)`

`2=A`

Thus,

`(13x)/((3x-2)(2x+3))=2/(3x-2)+3/(2x+3)`

So,

`int(13x)/(6x^2+5x-6)dx=2int1/(3x-2)dx+3int1/(2x+3)dx`

`=2/3int3/(3x-2)dx+3/2int2/(2x+3)dx`

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