Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 547: 9


$$\frac{4}{5} \ln 2+\frac{1}{5} \ln 3 $$

Work Step by Step

Given $$\int_{2}^{4}\frac{x+2}{x^2+3x-4}dx$$ Since \begin{align*} \frac{x+2}{x^2+3x-4}&=\frac{A}{x +4}+\frac{B}{x-1}\\ &= \frac{A(x-1)+B(x+4)}{(x +4)(x-1)}\\ x+2&=A(x-1)+B(x+4) \end{align*} At $x=1 \to B=3/5$ At $x=-4 \to B=2/5$ \begin{align*} \int_{2}^{4} \frac{x+2}{x^{2}+3 x-4} d x &=\int_{2}^{4}\left(\frac{2 / 5}{x+4}+\frac{3 / 5}{x-1}\right) d x\\ &=\left[\frac{2}{5} \ln |x+4|+\frac{3}{5} \ln |x-1|\right]_{2}^{4} \\ &=\left(\frac{2}{5} \ln 8+\frac{3}{5} \ln 3\right)-\left(\frac{2}{5} \ln 6+0\right)\\ &=\frac{4}{5} \ln 2+\frac{1}{5} \ln 3 \end{align*}
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