Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 29



Work Step by Step

$\int\frac{\sqrt{x}}{\sqrt{x}-3}dx$ $=\int\frac{\sqrt{x}-3+3}{\sqrt{x}-3}dx$ $=\int1$ $dx$ $+\int\frac{3}{\sqrt{x}-3}dx$ for $\int\frac{3}{\sqrt{x}-3}dx$, let u=$\sqrt{x}$ $du=\frac{1}{2\sqrt{x}}dx$ $du=\frac{1}{2u}dx$ $2u$ $du$ $=dx$ $\int\frac{3}{\sqrt{x}-3}dx$ $=3\int\frac{2u}{u-3}du$ $=6\int\frac{u}{u-3}fu$ $=6(\int1$ $du + 3\int\frac{1}{u-3}du)$ $=6u+18ln(u-3)$ $=6\sqrt{x}+18ln(\sqrt{x}-3)$ $\int1$ $dx$ $+\int\frac{3}{\sqrt{x}-3}dx$ $=x+6\sqrt{x}+18ln(\sqrt{x}-3)+C$
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