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: 65


$$\int \frac{dx}{2+(x-1)^2}=\frac{\sqrt2}{2}\tan^{-1}\Big(\frac{(x-1)\sqrt2}{2}\Big)+C$$

Work Step by Step

$$A=\int \frac{dx}{2+(x-1)^2}$$ We set $a=x-1$, which means $$da=dx$$ Therefore, $$A=\int\frac{da}{2+a^2}=\int\frac{da}{(\sqrt2)^2+a^2}$$ We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$$ So, $$A=\frac{1}{\sqrt2}\tan^{-1}\frac{a}{\sqrt2}+C$$ $$A=\frac{\sqrt2}{2}\tan^{-1}\frac{a\sqrt2}{2}+C$$ $$A=\frac{\sqrt2}{2}\tan^{-1}\Big(\frac{(x-1)\sqrt2}{2}\Big)+C$$
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