Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3: 17

Answer

$\frac{(7t^2+12)^\frac{3}{2}}{21}+C$

Work Step by Step

Use the substitution $u=7t^2+12$. Then $du=14tdt$, so $t dt=\frac{1}{14}du$. $\int t\sqrt{7t^2+12}dt$ $=\int \sqrt{u}*\frac{1}{14}du$ $=\int \frac{1}{14}\sqrt{u}du$ $=\frac{1}{14}*\frac{u^\frac{3}{2}}{\frac{3}{2}}+C$ $=\frac{u^\frac{3}{2}}{21}+C$ $=\frac{(7t^2+12)^\frac{3}{2}}{21}+C$
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