Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 328: 84


$ e^x(e^{2x}+1)^3- \frac{6}{7} e^{7x}- \frac{12}{5} e^{5x} -2e^{3x} +c $

Work Step by Step

We can do the integration by parts, $ u=(e^{2x}+1)^3,v'=e^xdx $, then $ u'=6e^{2x}(e^{2x}+1)^2dx $, $ v= e^x $. Hence, we have $$ \int e^x(e^{2x}+1)^3dx=e^x(e^{2x}+1)^3-6\int e^{3x}(e^{2x}+1)^2 dx\\\ =e^x(e^{2x}+1)^3-6\int e^{7x}+2e^{5x} +e^{3x} dx\\ =e^x(e^{2x}+1)^3- \frac{6}{7} e^{7x}- \frac{12}{5} e^{5x} -2e^{3x} +c . $$
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