Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.9 Representations of Functions as Power Series - 11.9 Exercises - Page 798: 27

Answer

Radius of convergence is $1$ $\int x^{2}ln(1+x)dx=c+\Sigma_{n=1}^{\infty}(-1)^{n}\dfrac{{x}^{n+3}}{n(n+3)}$

Work Step by Step

Sum of geometric series with initial term $a$ and common ratio $r$ is $S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$ Here, $x^{2}ln(1+x)=\frac{a}{1-r}$ we know that the power series converges when $r=|x|\lt 1$ Radius of convergence is $1$ $\int x^{2}ln(1+x)dx=c+\Sigma_{n=1}^{\infty}(-1)^{n}\dfrac{{x}^{n+3}}{n(n+3)}$
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