## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 7 - Practice Exercises - Page 420: 7

#### Answer

$$\int \frac{\ln(x-5)}{x-5}dx=\frac{\Big(\ln(x-5)\Big)^2}{2}+C$$

#### Work Step by Step

$$A=\int \frac{\ln(x-5)}{x-5}dx$$ Set $u=\ln(x-5)$, we have $$du=\frac{(x-5)'}{x-5}dx=\frac{1}{x-5}dx$$ Therefore, $$A=\int udu=\frac{u^2}{2}+C$$ $$A=\frac{\Big(\ln(x-5)\Big)^2}{2}+C$$

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