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: 55

Answer

$$ \frac{1}{4}x^4\ln^2 x-\frac{1}{2}\left(\frac{1}{4}x^4\ln x-\frac{1}{16}x^4\right)+C$$

Work Step by Step

Given $$\int x^{3} \ln ^{2} x d x$$ Let \begin{align*} u&=\ln ^2x\ \ \ \ \ \ \ \ \ \ dv=x^3dx\\ du&= \frac{2\ln x }{x}dx\ \ \ \ \ \ \ v = \frac{1}{4}x^4 \end{align*} Then \begin{align*} \int x^{3} \ln^2 x d x&= \frac{1}{4}x^4\ln^2 x-\frac{1}{2}\int x^3\ln xdx \end{align*} Let \begin{align*} u&=\ln x\ \ \ \ \ \ \ \ \ \ dv=x^3dx\\ du&= \frac{1 }{x}dx\ \ \ \ \ \ \ v = \frac{1}{4}x^4 \end{align*} Then \begin{align*} \int x^3\ln xdx &= \frac{1}{4}x^4\ln x-\frac{1}{4}\int x^3dx\\ &= \frac{1}{4}x^4\ln x-\frac{1}{16}x^4 \end{align*} Hence \begin{align*} \int x^{3} \ln^2 x d x&= \frac{1}{4}x^4\ln^2 x-\frac{1}{2}\left(\frac{1}{4}x^4\ln x-\frac{1}{16}x^4\right)+C \end{align*}
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