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: 23

Answer

$\ln|x-1|-\frac{1}{2}\ln|x^2+9|-\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)+C$

Work Step by Step

$\int\frac{10}{(x-1)(x^{2}+9)}dx$ = $\int\left(\frac{A}{x-1}+\frac{Bx+C}{x^2+9}\right)dx$ $10$ = $A(x^2+9)+(Bx+C)(x-1)$ $10$ = $(Ax^2+9A)+(Bx^2-Bx+Cx-C)$ $0$ = $A+B$ $0$ = $-B+C$ $10$ = $9A-C$ $A$ = $1$, $B$ = $-1$, $C$ = $-1$ So $\int\frac{10}{(x-1)(x^{2}+9)}dx$ = $\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)dx$ = $\int\left[\frac{1}{x-1}-\frac{x}{x^2+9}-\frac{1}{x^2+9}\right]dx$ = $\ln|x-1|-\frac{1}{2}\ln|x^2+9|-\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)+C$
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