Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 501: 32

Answer

$$\int_{0}^{1} \frac{x}{x^{2}+4 x+13} d x=\frac{1}{2} \ln \frac{18}{13}-\frac{\pi}{6}+\frac{2}{3} \tan ^{-1} \frac{2}{3}$$

Work Step by Step

Given $$\int_{0}^{1} \frac{x}{x^{2}+4 x+13} d x$$ So, we have \begin{aligned} I=& \int_{0}^{1} \frac{x}{x^{2}+4 x+13} d x \\=& \frac{1}{2} \int_{0}^{1} \frac{2 x}{x^{2}+4 x+13} d x \\=& \frac{1}{2} \int_{0}^{1} \frac{2 x+4}{x^{2}+4 x+13}-\frac{4}{x^{2}+4 x+13} d x \\=& \frac{1}{2} \int_{0}^{1} \frac{2 x+4}{x^{2}+4 x+13}-\frac{4}{(x+2)^{2}+9} d x \\=& \frac{1}{2} \int_{0}^{1} \frac{2 x+4}{x^{2}+4 x+13} d x-\frac{1}{2} \int_{0}^{1} \frac{4}{(x+2)^{2}+9} d x \\ &=\frac{1}{2} \left[\ln |x^2+4x+13|\right]_{0}^{1}-2\int_{0}^{1} \frac{1}{(x+2)^{2}+9} d x\end{aligned} Since $$\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$$ So, we get \begin{align} &I= \frac{1}{2} \left[\ln |x^2+4x+13|\right]_{0}^{1}-2\int_{0}^{1} \frac{1}{(x+2)^{2}+3^2} d x\\ &= \frac{1}{2} \ln 18-\frac{1}{2} \ln 13-\left[\frac{2}{3} \tan ^{-1}\left[\frac{x+2}{3}\right]\right]_{0}^{1} \\ & =\frac{1}{2} \ln \frac{18}{13}-\frac{2}{3}\left[\tan ^{-1}1-\tan ^{-1} \frac{2}{3}\right] \\ & =\frac{1}{2} \ln \frac{18}{13}-\frac{2}{3}\left[\frac{\pi}{4}-\tan ^{-1} \frac{2}{3}\right] \\ & =\frac{1}{2} \ln \frac{18}{13}-\frac{\pi}{6}+\frac{2}{3} \tan ^{-1} \frac{2}{3} \end{align}
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