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: 37


$$\ln \left|\frac{x+2}{3}+\frac{\sqrt{x^{2}+4 x+13}}{3}\right|+C $$ or $$ \ln \left|x+2+\sqrt{x^{2}+4 x+13}\right|+C $$

Work Step by Step

Given $$\int \frac{d x}{\sqrt{x^{2}+4 x+13}}$$ Since \begin{align*} x^{2}+4 x+13&= (x+2)^2-4+13\\ &=(x+2)^2-9 \end{align*} Let $$x+2=3\sec u\ \ \ \to \ \ \ \ dx=3\sec u\tan udu$$ Then \begin{align*} \int \frac{d x}{\sqrt{x^{2}+4 x+13}}&=\int \frac{d x}{\sqrt{(x+2)^2-9}}\\ &= \int \frac{3\sec u\tan udu}{\sqrt{9\sec^2 u-9}}\\ &= \int \frac{3\sec u\tan udu}{\sqrt{9\tan ^2u}}\\ &= \int \sec u du\\ &=\ln |\sec u+\tan u|+C\\ &= \ln \left|\frac{x+2}{3}+\frac{\sqrt{x^{2}+4 x+13}}{3}\right|+C \end{align*} or, we can write: $$ \ln \left|x+2+\sqrt{x^{2}+4 x+13}\right|+C $$
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