Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - Review Exercises - Page 395: 29


$\frac{1}{3}\ln {4.5}\approx.5014$

Work Step by Step

$\int\limits_2^3\frac{x^2+2x-2}{x^3+3x^2-6x}dx$ Now use the substitution rule for definite integrals $u=x^3+3x^2-6x$ $dx=\frac{du}{3x^2+6x-6}$ lower limit 2 goes to $(2)^3+3(2)^2-6(2)=8$ upper limit 3 goes to $(3)^3+3(3)^2-6(3)=36$ $\int\limits_8^{36}\frac{x^2+2x-2}{u}\frac{du}{3x^2+6x-6}=\int\limits_8^{36}\frac{1}{3}\frac{1}{u}du=\frac{1}{3}\int\limits_8^{36}\frac{1}{u}du$ $=\frac{1}{3}\ln u|_8^{36}$ (Fundamental theorem) $=\frac{1}{3}(\ln {36}-\ln 8)=\frac{1}{3}\ln {\frac{36}{8}}=\frac{1}{3}\ln {4.5}\approx.5014$
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