Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 58


$$\frac{x\ln x}{2} \sqrt{1+(x\ln x)^{2}}+\frac{1}{2} \ln (x\ln x+\sqrt{1+(x\ln x)^{2}})+C$$

Work Step by Step

Given $$\int(\ln x+1) \sqrt{(x \ln x)^{2}+1} d x$$ Let $$u=x\ln x \ \ \ \ \ \ \ \ \ \ du=\ln x+1 $$ Then $$\int(\ln x+1) \sqrt{(x \ln x)^{2}+1} d x=\int \sqrt{u^{2}+1} d u $$ Use $$ \int \sqrt{a^{2}+u^{2}} d u=\frac{u}{2} \sqrt{a^{2}+u^{2}}+\frac{a^{2}}{2} \ln (u+\sqrt{a^{2}+u^{2}})+C$$ Then \begin{align*} \int(\ln x+1) \sqrt{(x \ln x)^{2}+1} d x&=\int \sqrt{u^{2}+1} d u \\ &=\frac{u}{2} \sqrt{1+u^{2}}+\frac{1}{2} \ln (u+\sqrt{1+u^{2}})+C\\ &=\frac{x\ln x}{2} \sqrt{1+(x\ln x)^{2}}+\frac{1}{2} \ln (x\ln x+\sqrt{1+(x\ln x)^{2}})+C \end{align*}
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