Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 21

Answer

$-\frac 1 3 \ln|-t^3+9t+1|+C$

Work Step by Step

$Let u=-t^3+9t+1$ $du=(-3t^2+9)dt=-3(t^2-3)dt$ $\int \frac {t^2-3}{-t^3+9t+1}dt$ $-\frac 1 3 \int \frac {-3(t^2-3)}{-t^3+9t+1}dt$ $-\frac 1 3 \ln|-t^3+9t+1|+C$
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