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: 27


\[\int \frac{x^3+4x+3}{x^4+5x^2+4} dx =\frac{1}{2}\ln |x^2+1|+\arctan (x)-\frac{1}{2}\arctan (\frac{x}{2})+C\]

Work Step by Step

Make \[\frac{x^3+4x+3}{x^4+5x^2+4}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4} = \frac{(A+C)x^3+(B+D)x^2+(4A+C)x+4B+D}{x^4+5x^2+4} \] $\therefore$ \[ \begin{cases} A=1\\ B=1\\ C=0\\ D=-1\\ \end{cases} \] $\therefore$\[\int \frac{x^3+4x+3}{x^4+5x^2+4} dx = \int (\frac{x+1}{x^2+1}+\frac{-1}{x^2+4}) dx=\frac{1}{2}\ln |x^2+1|+\arctan (x)-\frac{1}{2}\arctan (\frac{x}{2})+C\]
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