Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 346: 19


$\frac{(x^3+3x)^5}{15} +C$

Work Step by Step

Evaluate the Integral using substitution: $\int (x^2+1)(x^3+3x)^4dx$ Substitution Rule: $\int f(g(x))gā€™(x)dx = \int f(u)du$ $u= x^3+3x$ $du =3x^2 + 3$ Since $du$ in the expression is equal to $(x^2+1)$ it must be multiplied by $\frac{1}{3}$ Solve the integral in terms of $u$: $\int (u)^4(\frac13)du$ $\frac{1}{3}\int (u)^4du $ $\frac{1}{3}\frac{u^5}{5} +C$ $\frac{u^5}{15} + C$ Substitute for $u$: $\frac{(x^3+3x)^5}{15} +C$
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