#\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#