   Chapter 11.9, Problem 27E

Chapter
Section
Textbook Problem

# Evaluate the indefinite integral as a power series. What is the radius of convergence?27. ∫ x 2 ln ( 1 + x )   d x

To determine

To evaluate: The indefinite integral as a power series and the radius of convergence

Explanation

Result used:

(1) “The sum of the geometric series with initial term a and common ratio r is S=n=0arn=a1r .”

(2)Ratio test:

“If limn|an+1an|=L<1 , then the series n=1an is absolutely convergent.”

Given:

The indefinite integral is x2ln(1+x)dx

Calculation:

Let f(t)=x2ln(1+x)dx

Modify the function as follows:

f(t)=[x211+xdx]dx

By the result (1) the function 11+x can be written as follows:

11+x=11(x)=a1r

Therefore, f(t) is a sum of a geometric series with initial term a=1 and r=x .

Show the expansion of series by using result (1).

f(t)=n=0arn=n=0(1)(x)n=n=0(1)nxn

f(x)=[x2n=0(1)nxndx]dx=[x2n=0(1)nxn+1n+1]dx

Therefore, the power series representation as follows:

Consider x2ln(1+x)=[x2n=0(1)nxn+1n+1]dx

Integrate both sides

x2ln(1+x)=[x2n=0(1)nxn+1n+1]dx=[n=0(1)nxn+3n+1]dx=[n=0(1)nxn+4(n+1)(n+4)]+C

Replace n by n+1 in the summation part

x2ln(1+x)=[n=1(1)nxn+3n(n+3)]+C

Therefore, the power series representation is

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