Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Practice Exercises - Page 517: 2


$\dfrac{x^3}{3}\ln x-(\dfrac{x^3}{9})+c$

Work Step by Step

The integration by parts formula suggests that $\int a'(x) b(x)=a(x) b(x)-\int a(x) b'(x)dx$ Now, we have $I=\int x^2 \ln x dx$ $\implies I=(\dfrac{x^3}{3})(\ln x)-\int (\dfrac{x^3}{3}) (\dfrac{1}{x})dx$ This implies that $I=(\dfrac{x^3}{3})(\ln x)-\int (\dfrac{x^2}{3})dx$ $\implies I=\dfrac{x^3}{3}\ln x-(\dfrac{x^3}{9})+c$
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