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: 20


$\dfrac{1}{2}e^{-\frac{1}{x^2}} +c$

Work Step by Step

Solve $\int \dfrac{e^{-1/x^2} }{x^3}dx$ Let us $p=-\dfrac{1}{x^2}$ and $dp=2(\dfrac{1}{x^2})dx$ This implies $\int \dfrac{e^{-1/x^2} }{x^3}dx=\dfrac{1}{2}\int e^p dp$ $=\dfrac{1}{2} e^p +c$ $=\dfrac{1}{2}e^{-\frac{1}{x^2}} +c$ Hence,$\int \dfrac{e^{-1/x^2} }{x^3}dx=\dfrac{1}{2}e^{-\frac{1}{x^2}} +c$
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