Answer
$-\frac 1 3 \ln|-t^3+9t+1|+C$
Work Step by Step
$Let u=-t^3+9t+1$
$du=(-3t^2+9)dt=-3(t^2-3)dt$
$\int \frac {t^2-3}{-t^3+9t+1}dt$
$-\frac 1 3 \int \frac {-3(t^2-3)}{-t^3+9t+1}dt$
$-\frac 1 3 \ln|-t^3+9t+1|+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 - Page 512: 21
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