## Calculus 10th Edition

Published by Brooks Cole

# Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 23

#### Answer

$-\frac{2}{3}$ $ln|(1-3\sqrt{x})|+C$

#### Work Step by Step

let u =$1-3\sqrt{x}$ du = $-\frac{3}{2}x^{-\frac{1}{2}}dx=-\frac{3}{2\sqrt{x}}dx$ $\int\frac{1}{\sqrt{x}(1-3\sqrt{x})}$ $=-\frac{2}{3}$ $\int\frac{1}{u}du$ $=-\frac{2}{3}ln|(u)|+C$ $=-\frac{2}{3}$ $ln|(1-3\sqrt{x})|+C$

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