Chapter 13 - Section 13.2 - Substitution - Exercises - Page 971: 54

$\displaystyle \frac{1}{a}\ln|ax+b|+C$

$\displaystyle \int(ax+b)^{-1}dx=\quad \left[\begin{array}{ll} u=ax+b, & du=adx, \\ & dx=\frac{1}{a}du \end{array}\right]$ $=\displaystyle \frac{1}{a}\int u^{-1}du$ apply power rule for $n=-1$ $=\displaystyle \frac{1}{a}\ln|u|+C$ bring back the variable $x$ $=\displaystyle \frac{1}{a}\ln|ax+b|+C$