Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 501: 24

Answer

$2\displaystyle \ln|x|-\frac{1}{2}\ln|x^{2}+3|-\frac{1}{\sqrt{3}}\arctan(\frac{x}{\sqrt{3}})+C$

Work Step by Step

$I= \displaystyle \int\frac{x^{2}-x+6}{x^{3}+3x}dx$ $ \displaystyle \frac{x^{2}-x+6}{x(x^{2}+3)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+3}$ $x^{2}-x+6=A(x^{2}+3)+x(Bx+C)$ $x^{2}-x+6=Ax^{2}+3A+Bx^{2}+Cx$ $x^{2}-x+6=(A+B)x^{2}+Cx+3A$ $3A=6\Rightarrow A=2$ $C=-1$ $A+B=1\Rightarrow B=-1$ $ \displaystyle \frac{x^{2}-x+6}{x(x^{2}+3)}=\frac{2}{x}+\frac{-2x-1}{x^{2}+3}=\frac{2}{x}-\frac{2x}{x^{2}+3}-\frac{1}{x^{2}+(\sqrt{3})^{2}}$ $I=2\displaystyle \ln|x|-\frac{1}{2}\ln|x^{2}+3|-\frac{1}{\sqrt{3}}\arctan(\frac{x}{\sqrt{3}})+C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.