Integrating Using Partial Fractions

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Homework Help Overview

This discussion revolves around an arc length problem in three dimensions involving the vector function r(t) = . Participants are exploring the integration process, particularly focusing on the use of partial fractions after simplifying the integral.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to compute the arc length by taking the derivative of the vector and simplifying the resulting integral. Some participants suggest additional substitutions to simplify the expression further before applying partial fractions. Others express challenges with the approach, indicating that their attempts to split the expression have led to complications.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and experiences. Some guidance has been offered regarding substitution techniques and the structure of the expression to avoid complications like long division. There is no explicit consensus on the best approach yet, as participants are exploring different methods.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may impose specific methods or techniques to be used, such as the requirement to apply partial fractions. There is also a noted complexity in the expressions being manipulated, which may affect the clarity of the discussion.

Jay9313
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Homework Statement


This is an arc length problem in three dimensions. I was given the vector r(t)=<et, 1, t> from t=0 to t=1


Homework Equations


Arc Length= \int |\sqrt{r&#039;(t)}| dt from t1 to t2
where |\sqrt{r&#039;(t)}| is the magnitude of the derivative of the vector

The Attempt at a Solution



I took the derivative and got the magnitude and simplified it down to
∫ √(e2t+1) dt
I then set u=e2t+1
I then simplified and substituted until I got to:∫\frac{\sqrt{(u)}}{u-1} du

My professor said to use partial fractions from here, but I'm not sure how to do that.
 
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Try one more substitution to get rid of the square root on top first. Then do partial fractions.
 
That's not really working at all
 
What did you try?
 
I split it up, and it made it worse, and I had to do long division, but I got an answer. It's nasty, but I got an answer.
 
Presumably, you ended up with something like ##\frac{v^2}{v^2-1}## after the substitution. A good technique to avoid doing long division is to add and subtract judiciously:
$$\frac{v^2}{v^2-1} = \frac{(v^2-1)+1}{v^2-1} = 1 + \frac{1}{v^2-1}.$$
 

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