Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 5 - Accumulating Change: Limits of Sums and the Definite Integral - Review Activities - Page 415: 22


$$\frac{1}{3}e^{3 x}-\frac{\left( 5 \right)^{-x}}{\ln 5}+x\ln (1.4x)- x+c$$

Work Step by Step

Given $$ \int\left[e^{3 x}+\ln (1.4 x)+\left(\frac{1}{5}\right)^{x}\right] d x $$ Since \begin{align*} \int\left[e^{3 x}+\ln (1.4 x)+\left(\frac{1}{5}\right)^{x}\right] d x&= \int\left[e^{3 x}+\ln (1.4 x)+\left( 5 \right)^{-x}\right] d x\\ &=\frac{1}{3}e^{3 x}-\frac{\left( 5 \right)^{-x}}{\ln 5}+\int\ln (1.4 x)d x \end{align*} To evaluate $ \int\ln (1.4 x)d x$, let $u=1.4x\ \to\ du=1.4dx$ and apply the rule $$ \int \ln u d u=u \ln u-u+c $$ We get \begin{align*} \ln (1.4x)dx&=\frac{1}{1.4}\int\ln udu\\ &=\frac{1}{1.4}\left(u\ln u-u\right)+c\\ &=\frac{1}{1.4}\left(1.4x\ln (1.4x)-1.4x\right)+c\\ &=x\ln (1.4x)- x+c \end{align*} Hence $$\int\left[e^{3 x}+\ln (1.4 x)+\left(\frac{1}{5}\right)^{x}\right] d x =\frac{1}{3}e^{3 x}-\frac{\left( 5 \right)^{-x}}{\ln 5}+x\ln (1.4x)- x+c$$
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