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: 34

Answer

$\int \frac{x^5+ x - 1}{x^3 +1} dx $ $=$ $\frac{x^3}{3} - ln | x+1| +c$

Work Step by Step

$ \frac{x^5+ x - 1}{x^3 +1} $ $= x^2 + \frac{-1(x^2-x+1)}{x^3 +1}$ $Thus ,$ $\int x^2 +\frac{-1(x^2-x+1)}{x^3 +1} dx $ $=\int x^2 +\frac{-1(x^2-x+1)}{(x +1)(x^2 -x+1) } dx $ $=\int x^2 +\frac{-1}{x +1} dx $ $= \frac{x^3}{3} - ln |x+1| +c$
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