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: 11


$\dfrac{1}{8}(\ln 4)^4 +c$

Work Step by Step

Solve $\int_1^4\dfrac{(\ln x)^3}{2x}dx$ Let us $p= \ln x$ and $dp=\dfrac{1}{x}dx$ This implies $\int_1^4\dfrac{(\ln x)^3}{2x}dx= \dfrac{1}{2} \int_1^4 u^3 du$ $= \dfrac{1}{8} [u^4]_1^4+c$ $= \dfrac{1}{8}(\ln 4)^4 +c$ Hence, $\int_1^4\dfrac{(\ln x)^3}{2x}dx=\dfrac{1}{8}(\ln 4)^4 +c$
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