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: 16

Answer

$= -\frac{z}{\ln 10\times10^{z}} - \frac{1}{(\ln 10)^{2}\times 10^{z}} + C$

Work Step by Step

$\int \frac{z}{10^{z}}dz$ $\int z(10)^{-z}dz$ $u = z$ $u' = 1$ $\frac{du}{dz} = 1$ $du = 1 dz$ $dv = 10^{-z}dz$ $dv = e^{\ln (10^{-z})}$ $dv = e^{-z \ln 10}$ $v = \frac{1}{-\ln 10}e^{-z \ln 10}$ $v = \frac{1}{-\ln 10}10^{-z}$ $v = -\frac{1}{\ln 10\times10^{z}}$ $uv - \int vdu$ $= (z)(-\frac{1}{\ln 10\times10^{z}}) - \int (-\frac{1}{\ln 10\times10^{z}}) dz$ $= (z)(-\frac{1}{\ln 10\times10^{z}}) - \int \frac{1}{-\ln 10}10^{-z} dz$ $= (z)(-\frac{1}{\ln 10\times10^{z}}) + \frac{1}{\ln 10}\int 10^{-z} dz$ $= (z)(-\frac{1}{\ln 10\times10^{z}}) - \frac{1}{\ln 10}(\frac{1}{\ln 10})10^{-z} + C$ $= (z)(-\frac{1}{\ln 10\times10^{z}}) - \frac{1}{(\ln 10)^{2}}10^{-z} + C$ $= -\frac{z}{\ln 10\times10^{z}} - \frac{1}{(\ln 10)^{2}\times 10^{z}} + C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.