Answer
$$\frac{1}{4}\sqrt{4x^2+9}+c .$$
Work Step by Step
Let $u=4x^2+9$, then $du=8xdx$. Now, we have
$$\int \frac{xdx}{\sqrt{4x^2+9}}=\frac{1}{8}\int \frac{du}{\sqrt u}=\frac{1}{8} \frac{\sqrt u}{1/2}+c\\
=\frac{1}{4}\sqrt{4x^2+9}+c .$$
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 - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 75
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