How Do You Evaluate the Indefinite Integral as a Power Series?

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The discussion focuses on evaluating the indefinite integral ∫[ln(1−t)/7t]dt as a power series. The first five non-zero terms of the power series representation centered at t=0 can be derived by expanding ln(1−t) using the Maclaurin series. The radius of convergence for this power series is established as R=1. Participants emphasize the importance of including a constant "C" in the final expression.

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Evaluate the indefinite integral as a power series
∫[ln(1−t)/7t]dt.
Find the first five non-zero terms of power series representation centered at t=0.

Answer: f(t)=

What is the radius of convergence?
Answer: R= 1

Note: Remember to include a constant "C".

This problem has been difficult for me for awhile now. I have my radius but I can't figure out my five f(t) terms it's asking for.
Try to build those terms from this:

(-t^(n))/(7n^(2))

Also n=1
 
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ajkess1994 said:
Evaluate the indefinite integral as a power series
∫[ln(1−t)/7t]dt.
Find the first five non-zero terms of power series representation centered at t=0.

Answer: f(t)=

What is the radius of convergence?
Answer: R= 1

Note: Remember to include a constant "C".

This problem has been difficult for me for awhile now. I have my radius but I can't figure out my five f(t) terms it's asking for.
Try to build those terms from this:

(-t^(n))/(7n^(2))

Also n=1

Hi ajkess1994,

Can you expand $\ln(1−t)$ as a MacLauren series?
Substitute it, and the rest should follow.
 
Yes, Arsen I can expan the ln(1-t) to a MacLauren series, and from there finding the "nth" terms shouldn't be too difficult, thank you for the suggestioon.
 

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