University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 16


$-2 e^{-\sqrt r} +c$

Work Step by Step

Solve $\int e^{-\sqrt r} \dfrac{1}{\sqrt r}dr$ Let us $p=-\sqrt r$ and $dp=-\dfrac{1}{2 \sqrt r}dr$ This implies $\int e^{-\sqrt r} \dfrac{1}{\sqrt r}dr=-2\int e^p dp $ $=-2 e^{p} +c$ $=-2 e^{-\sqrt r} +c$ Hence, $\int e^{-\sqrt r} \dfrac{1}{\sqrt r}dr=-2 e^{-\sqrt r} +c$
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