Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.4 Integration of Rational Functions by Partial Fractions - 7.4 Exercises - Page 541: 33


$$\int_{0}^{1} \frac{x^{3}+2 x}{x^{4}+4 x^{2}+3} d x=\frac{1}{4} \ln \frac{8}{3}$$

Work Step by Step

Given $$\int_{0}^{1} \frac{x^{3}+2 x}{x^{4}+4 x^{2}+3} d x$$ \begin{array}{l}{\text { Let } u=x^{4}+4 x^{2}+3, \text { so that } d u=\left(4 x^{3}+8 x\right) d x}\\{du=4\left(x^{3}+2 x\right) d x, x=0 \Rightarrow u=3, \text { and } x=1 \Rightarrow u=8} \\ {\text { Then } }\\{\begin{align}I&=\int_{0}^{1} \frac{x^{3}+2 x}{x^{4}+4 x^{2}+3} d x\\&=\int_{3}^{8} \frac{1}{u}\left(\frac{1}{4} d u\right)\\ &=\frac{1}{4}[\ln |u|]_{3}^{8}\\&=\frac{1}{4}(\ln 8-\ln 3)\\ &=\frac{1}{4} \ln \frac{8}{3}\end{align}}\end{array}
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