Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 35


$$ \ln |\sqrt{(x-2)^{2}+4}+x-2|+C$$

Work Step by Step

Given $$\int \frac{d x}{\sqrt{x^{2}-4 x+8}}$$ Since $$ x^{2}-4 x+8= (x-2)^{2}+4$$ Let $$u=x-2\ \ \ \ \to\ \ \ du=dx $$ Then $$\int \frac{d x}{\sqrt{x^{2}-4 x+8}}=\int \frac{d u}{\sqrt{u^{2}+2^{2}}}$$ Let $$ u=2\tan z\ \ \ \ du=2\sec^2 zdz$$ Then \begin{align*} \int \frac{d x}{\sqrt{x^{2}-4 x+8}}&=\int \frac{d u}{\sqrt{u^{2}+2^{2}}}\\ &=\int \frac{2\sec^2 zdz }{\sqrt{4+4\tan^2 z}}\\ &= \int \sec zdz\\ &= \ln |\sec z+\tan z|+C\\ &=\ln \left|\frac{\sqrt{u^{2}+4}}{2}+\frac{u}{2}\right|+C_{1}\\ &=\ln |\sqrt{u^{2}+4}+u|+\left(\ln \frac{1}{2}+C_{1}\right)=\ln |\sqrt{u^{2}+4}+u|+C\\ &= \ln |\sqrt{(x-2)^{2}+4}+x-2|+C \end{align*}
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