Integration Problem

natHenderson

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Joined
May 17, 2020
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32
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So far I've tried u-substitution and set the u = e^x - 1. This gave me the integral of 1/(1+u^2)du which is arctan(u) or arctan(e^x-1), then i set the pike from 0 to ln(2), but that just gave me arctan(3) - arctan(2) which isn't one of the answers, does anybody know what I did wrong?
 

Subhotosh Khan

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Jun 18, 2007
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What did you get when you set the limits? - show detailed work. Your integration is correct. We would not know "what you did wrong" - unless you show us your work regarding the limits.
 

topsquark

Full Member
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Aug 27, 2012
Messages
920
You got the limits wrong.
\(\displaystyle tan^{-1} \left ( e^x - 1 \right )_{x = ln(2)} = tan^{-1} ( 2 - 1) = tan^{-1}(1) = \dfrac{ \pi }{4}\)

Your lower limit is off, too.

-Dan
 

natHenderson

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May 17, 2020
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32
Oh ok, thanks Dan!
 
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