Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 547: 14

Answer

$$x\ln(x^2+1)-2x+2\tan^{-1}x+c$$

Work Step by Step

Given $$\int \ln(x^2+1)dx$$ Let \begin{align*} u&= \ln(x^2+1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=dx\\ du&= \frac{2x}{x^2+1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v= x \end{align*} Then \begin{align*} \int \ln(x^2+1)dx&=x\ln(x^2+1)-\int \frac{2x^2dx}{x^2+1} \\ &=x\ln(x^2+1)-2\int \frac{ x^2 +1-1 }{x^2+1}dx\\ &=x\ln(x^2+1)-2\int \left(1-\frac{ 1 }{x^2+1}\right)dx\\ &=x\ln(x^2+1)-2x+2\tan^{-1}x+c \end{align*}
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