Question #b6028

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Answer 1 · 2018-01-05T01:52:27

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

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:

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