\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! ☺