\int1/(x-b)dx?

Apparently I have to do this via partial fractions.
I know the answer is \ln|x-b|+C but I don't know how to achieve this using partial fractions.

I can do it with u-substitution, though --
\int1/(x-b)dx\rArr (u=x-b, du=1dx) \rArr\int1/udu=\ln|u|=\ln|x-b|+C