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#