Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 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

$\int t^{4} \ln(t) dt$ $u = \ln t$ $u' = \frac{1}{t}$ $\frac{du}{dt} = \frac{1}{t}$ $du = \frac{dt}{t}$ $dv = t^{4}$ $v = \frac{t^{5}}{5}$ $uv - \int vdu$ $= (\ln t)(\frac{t^{5}}{5}) - \int \frac{t^{5}}{5} \frac{dt}{t}$ $= (\ln t)(\frac{t^{5}}{5}) - \int \frac{t^{5}}{5} \frac{1}{t}dt$ $= (\ln t)(\frac{t^{5}}{5}) - \int \frac{t^{5}}{5t}dt$ $= (\ln t)(\frac{t^{5}}{5}) - \int \frac{t^{4}}{5}dt$ $= (\ln t)(\frac{t^{5}}{5}) - \frac{1}{5}\int t^{4}dt$ $= (\ln t)(\frac{t^{5}}{5}) - \frac{1}{5}\frac{t^{5}}{5}$ $= (\ln t)(\frac{t^{5}}{5}) - \frac{t^{5}}{25} +C$ $= (\frac{1}{5})t^{5}(\ln t) - \frac{1}{25}t^{5} +C$
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