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: 17

Answer

$\frac{9}{5}\ln{\frac{8}{3}}$

Work Step by Step

$\int_1^2{\frac{4y^{2}-7y-12}{y(y+2)(y-3)}}dy$ = $\int_1^2\left({\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}}\right)dy$ $4y^{2}-7y-12$ = $A(y+2)(y-3)+By(y-3)+Cy(y+2)$ $4y^{2}-7y-12$ = $A(y^{2}-y-6)+B(y^{2}-3y)+C(y^{2}+2y)$ $$\begin{cases} A+B+C& = 4\\ -A-3B+2C&= -7\\ -6A& = -12. \end{cases}$$ $A = 2$, $B = \frac{9}{5}$, $C = \frac{1}{5}$ So $\int_1^2{\frac{4y^{2}-7y-12}{y(y+2)(y-3)}}dy$ = $\int_1^2\left[{\frac{2}{y}+\frac{9}{5(y+2)}+\frac{1}{5(y-3)}}\right]dy$ = $\left[2\ln{y}+\frac{9}{5}{\ln|y+2|}+\frac{1}{5}{\ln|y-3|}\right]_1^2$ = $(2\ln2+\frac{9}{5}\ln4+\frac{1}{5}\ln1)-(2\ln1+\frac{9}{5}\ln3+\frac{1}{5}\ln2)$ = $\frac{9}{5}\ln{\frac{8}{3}}$
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