Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 5 - Accumulating Change: Limits of Sums and the Definite Integral - 5.9 Activities - Page 409: 14


$$\frac{1}{2}\left(\left(x^{2}+1\right) \ln \left(x^{2}+1\right)-\left(x^{2}+1\right)\right)+c$$

Work Step by Step

Given: $$\int x \ln \left(x^{2}+1\right) d x$$ Let $u=x^2+1\ \to\ \ du=2xdx$, so: \begin{align*} \int x \ln \left(x^{2}+1\right) d x&=\frac{1}{2}\int \ln \left(u\right) du \end{align*} Apply the Natural Log Rule for antiderivatives: $$ \int \ln u d u=u \ln u-u+c $$ Thus: \begin{align*} \int x \ln \left(x^{2}+1\right) d x&=\frac{1}{2}\int \ln \left(u\right) du \\ &=\frac{1}{2}\left(u \ln u-u+\right)c\\ &=\frac{1}{2}\left(\left(x^{2}+1\right) \ln \left(x^{2}+1\right)-\left(x^{2}+1\right)\right)+c \end{align*}
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