## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 7 - Integration - Chapter Review - Review Exercises - Page 417: 31

#### Answer

$$\frac{1}{6}{e^{3{x^2}}} + C$$

#### Work Step by Step

\eqalign{ & \int {x{e^{3{x^2}}}} dx \cr & {\text{set }}u = 3{x^2}{\text{ then }}\frac{{du}}{{dx}} = 6x,\,\,\,\,\,\,\frac{1}{6}du = xdx \cr & {\text{write the integrand in terms of }}u \cr & \int {x{e^{3{x^2}}}} dx = \int {{e^u}} \left( {\frac{1}{6}du} \right) \cr & = \frac{1}{6}\int {{e^u}} du \cr & {\text{integrating by the formula }}\int {{e^u}} du = {e^u} + C \cr & = \frac{1}{6}{e^u} + C \cr & {\text{replace }}3{x^2}{\text{ for }}u \cr & = \frac{1}{6}{e^{3{x^2}}} + C \cr}

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