Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 29


$$ \frac{1}{3}x^3 \ln x-\frac{1}{9} x^3 +C $$

Work Step by Step

Given $$\int x^{2} \ln x d x$$ Let \begin{align*} u&= \ln x\ \ \ \ \ \ dv= x^2 d x\\ du&=\frac{1}{x} dx\ \ \ \ \ \ v=\frac{1}{3}x^3 \end{align*} Then \begin{align*} \int x^{2} \ln x d x&=\frac{1}{3}x^3 \ln x-\frac{1}{3}\int x^2 dx\\ &= \frac{1}{3}x^3 \ln x-\frac{1}{9} x^3 +C \end{align*}
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