Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 4

$\ln|x-5|$ + C

$\int\frac{1}{x-5} dx $ Let $u = x-5$ $\frac{du}{dx}$ = 1 $\frac{du}{1}$ = $dx$ Substitute $u$ and $dx$ into the original equation $\int\frac{1}{u}\frac{du}{1}$ = $\int\frac{1}{u} du$ = $\ln|u|$ + C Since $u = x-5$, substituting it back will give you $\ln|x-5|$ + C