Answer
$$\int\frac{1}{\sqrt{9+x^2}}dx= \sinh^{-1}(x/3)+c.$$
Work Step by Step
Since $\frac{d}{dx} \sinh^{-1}x=\frac{1}{\sqrt{x^2+1}}$, then we have
$$\int\frac{1}{\sqrt{9+x^2}}dx=\int\frac{1}{3\sqrt{1+(x/3)^2}}dx=\sinh^{-1}(x/3)+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 55
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