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: 67


$$\int \frac{dx}{(2x-1)\sqrt{(2x-1)^2-4}}=\frac{1}{4}\sec^{-1}\Big|\frac{2x-1}{2}\Big|+C$$

Work Step by Step

$$A=\int \frac{dx}{(2x-1)\sqrt{(2x-1)^2-4}}$$ We set $a=2x-1$, which means $$da=2dx$$ $$dx=\frac{1}{2}da$$ Therefore, $$A=\frac{1}{2}\int\frac{da}{a\sqrt{a^2-2^2}}$$ We have $$\int\frac{dx}{x\sqrt{x^2-u^2}}=\frac{1}{u}\sec^{-1}\Big|\frac{x}{u}\Big|+C$$ So, $$A=\frac{1}{2}\times\Big(\frac{1}{2}\sec^{-1}\Big|\frac{a}{2}\Big|\Big)+C$$ $$A=\frac{1}{4}\sec^{-1}\Big|\frac{a}{2}\Big|+C$$ $$A=\frac{1}{4}\sec^{-1}\Big|\frac{2x-1}{2}\Big|+C$$
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