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: 12

Answer

$\ln{\frac{3}{8}}$

Work Step by Step

Factor the denominator: $\int_0^1{\frac{x-4}{x^{2}-5x+6}}dx$ = $\int_0^1{\frac{x-4}{(x-2)(x-3)}}dx$ Write the fraction as a sum of two fractions: $\frac{x-4}{(x-2)(x-3)}$ = $\frac{A}{x-2}+\frac{B}{x-3}$ Determine $A$ and $B$: $x-4$ = $A(x-3)+B(x-2)$ $$\begin{align*} \begin{cases} A+B &= 1\\ -3A-2B &= -4. \end{cases} \end{align*}$$ $A = 2$, $B = -1$ So $\int_0^1{\frac{x-4}{x^{2}-5x+6}}dx$ = $\int_0^1\left({\frac{2}{x-2}+\frac{-1}{x-3}}\right)dx$ $=[2\ln|x-2|-\ln|x-3|]_0^1$ $=(2\ln1-\ln2)-(2\ln2-\ln3)$ = $-3\ln2+\ln3$ $=\ln{\frac{3}{8}}$
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