University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

Chapter 5 - Practice Exercises - Page 343: 66

Answer

$$\int \frac{dx}{1+(3x+1)^2}=\frac{1}{3}\tan^{-1}(3x+1)+C$$

Work Step by Step

$$A=\int \frac{dx}{1+(3x+1)^2}$$ We set $a=3x+1$, which means $$da=3dx$$ $$dx=\frac{1}{3}da$$ Therefore, $$A=\frac{1}{3}\int\frac{da}{1+a^2}$$ We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$$ So, $$A=\frac{1}{3}\times\frac{1}{1}\tan^{-1}\frac{a}{1}+C$$ $$A=\frac{1}{3}\tan^{-1}a+C$$ $$A=\frac{1}{3}\tan^{-1}(3x+1)+C$$

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