Question

Question #b6028

Answer 1

`3ln|(x-2)/(x-1)|+C`

Explanation:

We have that:

`int 3/((x-1)(x-2))dx`

the best way to proceed would be by partial fractions; so we should split the quantity to be integrated like so:

`3/((x-1)(x-2))=A/(x-1)+B/(x-2)`

Now multiplying this equation through by the denominator `(x-1)(x-2)`:

`-> 3=A(x-2)+B(x-1)`

Let `x=2` to cancel the A term:

`->3=B`

And also let `x=1` to cancel the B term:

`3=-A -> A=-3` so we now have values for `A` and `B` and so it follows that:

`int 3/((x-1)(x-2))dx=int3/(x-2)-3/(x-1)dx`

The integral can now be evaluated straightforwardly as:

`=3ln|x-2|-3ln|x-1|+C = 3ln|(x-2)/(x-1)|+C`