Answer
$-\frac{8^{-x}}{\ln 8} +C$
Work Step by Step
Let $\log x$ denote the natural logarithm $\ln x$.
$\int 8^{-x}$ $dx$
let $u=-x$
$du=-dx$
$\int 8^{-x}$ $dx$
$=- \int 8^u$ $du$
$=-\frac{8^u}{\log 8} +C$
$=-\frac{8^{-x}}{\log 8} +C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 363: 72
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