Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.1 Integration by Parts - 7.1 Exercises: 11

Answer

$\frac{1}{5}t^5 \ln t- \frac{1}{25}t^5 +C$

Work Step by Step

Let $u = ln t, dv = t^4 dt$ and $du = \frac{1}{t} dt, v = \frac{1}{5}t^5$ Then $\int t^4 \ln t dt$ $=\frac{1}{5} t^5 \ln t -\int \frac{1}{5} t^5 \frac{1}{t} dt$ $=\frac{1}{5} t^5 \ln t -\int \frac{1}{5} t^4 dt$ $=\frac{1}{5}t^5 \ln t- \frac{1}{25}t^5 +C$
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