Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - Review - Exercises - Page 578: 31

Answer

$\frac{12-3\pi }{2}$

Work Step by Step

Given $$\int_0^{\ln 10} \frac{e^x \sqrt{e^x-1}}{e^x+8} d x $$ Let $$u^2=e^x-1\ \ \to \ \ 2 u d u=e^x d x$$ At $x=0\ \ \to \ u= 0,\ \ \ x=\ln (10)\to \ \ u =3$, then \begin{aligned} \int_0^{\ln 10} \frac{e^x \sqrt{e^x-1}}{e^x+8} d x & =\int_0^3 \frac{u \cdot 2 u d u}{u^2+9}\\ &=2 \int_0^3 \frac{u^2}{u^2+9} d u\\ &=2 \int_0^3\left(1-\frac{9}{u^2+9}\right) d u \\ & =2\left[u-\frac{9}{3} \tan ^{-1}\left(\frac{u}{3}\right)\right]_0^3\\ &=2\left[\left(3-3 \tan ^{-1} 1\right)-0\right]\\ &=2\left(3-3 \cdot \frac{\pi}{4}\right)\\ &=6-\frac{3 \pi}{2}\\ &=\frac{12-3\pi }{2} \end{aligned}
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