Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 347: 67



Work Step by Step

To evaluate the integral $$\int\frac{dx}{5-3x}$$ we will use substitution $5-3x=t$ which gives us $-3dx=dt\Rightarrow dx=-\frac{dt}{3}$. Putting this into the integral we get: $$\int\frac{dx}{5-3x}=\int\frac{1}{t}\Big(-\frac{dt}{3}\Big)=-\frac{1}{3}\int \frac{1}{t}dt=-\frac{1}{3}\ln|t|+c$$ where $c$ is arbitrary constant. Now we have to express our solution in terms of $x$: $$\int\frac{dx}{5-3x}=-\frac{1}{3}\ln|t|+c=-\frac{1}{3}\ln|5-3x|+c$$
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