Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.6 - Integral Tables and Computer Algebra Systems - Exercises 8.6 - Page 481: 38



Work Step by Step

Let $I=\int\dfrac{x^2}{\sqrt{(x-2)^2+1}}dx$ Suppose $u=x-2 \implies du=dx$ and $x=u+2 \implies x^2=u^2+4u+4$ Therefore, $I=\int\frac{u^2+4u+4}{\sqrt{u^2+1}}du= \int\dfrac{u^2}{\sqrt{u^2+1}}du+4\int\dfrac{u}{\sqrt{u^2+1}}du+4\int\dfrac{du}{\sqrt{u^2+1}}$ Now, $I=-\dfrac{1}{2}\ln(u+\sqrt{u^2+1})+\dfrac{u\sqrt{u^2+1}}{2}+4\int da+4\sinh^{-1}u+C \\=-\dfrac{1}{2}\ln(u+\sqrt{u^2+1})+\dfrac{u\sqrt{u^2+1}}{2}+4a+4\sinh^{-1}u+C \\ =-\dfrac{1}{2}\ln(x-2+\sqrt{(x-2)^2+1})+\dfrac{(x-2)\sqrt{(x-2)^2+1}}{2}+4\sqrt{(x-2)^2+1}+4 \space \sinh^{-1}(x-2)+C$
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