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: 40


$$\frac{3}{2}\left[\frac{(x-2) \sqrt{x^{2}-4 x+7}}{3}+\ln \left|\frac{(x-2)+\sqrt{x^{2}-4 x+7}}{\sqrt{3}}\right|\right]+C$$

Work Step by Step

Given $$\int \sqrt{x^{2}-4 x+7} d x$$ Since \begin{align*} x^{2}-4 x+7&=(x-2)^{2}-4 +7\\ &=(x-2)^{2}+3 \end{align*} Let $$x-2=\sqrt{3}\tan u\ \ \ \ \ \ dx=\sqrt{3}\sec^2 udu $$ then \begin{align*} \int \sqrt{x^{2}-4 x+7} d x&=\int \sqrt{(x-2)^{2}+3} d x\\ &=\sqrt{3}\int \sqrt{3\tan^{2}u+3} \sec^2 udu \\\\ &=3\int \sec^3udu \end{align*} Use $$\int \sec ^{n} u d u=\frac{1}{n-1} \tan u \sec ^{n-2} u+\frac{n-2}{n-1} \int \sec ^{n-2} u d u $$ then $$\int \sec ^{3} u d u=\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+C$$ Hence \begin{align*} \int \sqrt{x^{2}-4 x+7} d x &=3\int \sec^3udu\\ &= \frac{3}{2} \sec u \tan u+\frac{3}{2} \ln |\sec u+\tan u|+C\\ &=\frac{3}{2}\left[\frac{u \sqrt{u^{2}+3}}{3}+\ln \left|\frac{u+\sqrt{u^{2}+3}}{\sqrt{3}}\right|\right]+C\\ &=\frac{3}{2}\left[\frac{(x-2) \sqrt{x^{2}-4 x+7}}{3}+\ln \left|\frac{(x-2)+\sqrt{x^{2}-4 x+7}}{\sqrt{3}}\right|\right]+C \end{align*}
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