# \int1/(x-b)dx?

## Apparently I have to do this via partial fractions. I know the answer is $\setminus \ln | x - b | + C$ but I don't know how to achieve this using partial fractions. I can do it with u-substitution, though -- $\setminus \int \frac{1}{x - b} \mathrm{dx} \setminus \Rightarrow$ ($u = x - b$, $\mathrm{du} = 1 \mathrm{dx}$) $\setminus \Rightarrow \setminus \int \frac{1}{u} \mathrm{du} = \setminus \ln | u | = \setminus \ln | x - b | + C$

$\setminus \ln | x - b | + C$