Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.1 Integration by Parts - 7.1 Exercises: 12

Answer

$y \tan^{-1}2y - \frac{1}{4} \ln(1+4y^2) + C$

Work Step by Step

Let $u = \tan^{-1} 2y, dv = dy$ and $du = \frac{2}{1+4y^2} dy, v = y$ Then $\int \tan^{-1}2y dy$ $= y \tan^{-1} 2y - \int \frac{2y}{1+4y^2} dy$ $= y \tan^{-1} 2y - \int \frac{1}{t} (\frac{1}{4} dt)$ $= y \tan^{-1} 2y - \frac{1}{4}\ln |t| + C$ $= y \tan^{-1} 2y - \frac{1}{4}\ln (1+4y^2) + C$
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