#\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#
Apparently I have to do this via partial fractions.
I know the answer is
I can do it with u-substitution, though --
1 Answer
Apr 3, 2018
Explanation:
Explained above (in question content). Thanks to Steve M for the check! ☺
