University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 68


$$\int \frac{dx}{(x+3)\sqrt{(x+3)^2-25}}=\frac{1}{5}\sec^{-1}\Big(\Big|\frac{x+3}{5}\Big|\Big)+C$$

Work Step by Step

$$A=\int \frac{dx}{(x+3)\sqrt{(x+3)^2-25}}$$ We set $a=x+3$, which means $$da=dx$$ Therefore, $$A=\int\frac{da}{a\sqrt{a^2-25}}=\int\frac{da}{a\sqrt{a^2-5^2}}$$ We have $$\int\frac{dx}{x\sqrt{x^2-u^2}}=\frac{1}{u}\sec^{-1}\Big|\frac{x}{u}\Big|+C$$ So, $$A=\frac{1}{5}\sec^{-1}\Big|\frac{a}{5}\Big|+C$$ $$A=\frac{1}{5}\sec^{-1}\Big(\Big|\frac{x+3}{5}\Big|\Big)+C$$
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