Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 5

Answer

$$ \frac{x^4}{16}(4\ln x -1).$$

Work Step by Step

We have $u=\ln x$ and $dv=x^3dx$. Then, $du=\frac{dx}{x}$, $v=\frac{x^4}{4}$ $$ \int x^3 \ln x dx=\int udv=uv-\int vdu\\ = \frac{x^4\ln x}{4}- \int \frac{x^3}{4} dx\\ =\frac{x^4\ln x}{4}- \frac{x^4}{16} \\ = \frac{x^4}{16}(4\ln x -1).$$
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