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BECCALICIOUS! Offline
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Calc 2 Integral Problem. - February 16th 2012, 04:09 PM

I had this problem on my test today and I wasn't really sure how to solve it.

∫ dx / (1 + (e^x) )

Anyone have any ideas?


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Re: Calc 2 Integral Problem. - February 16th 2012, 06:17 PM

You have to do it by parts or substitution with u and du.
u=e^x
du=e^x dx
= ∫1/(u (u+1)) du
∫(1/u-1/(u+1)) du
Integrate and factor...
= ∫1/u du- ∫1/(u+1) du
If you intergrate 1/(u+1) it gives you log(u+1)
= ∫1/u du-log(u+1)
Integrate 1/u and it gives you log(u)
= log(u)-log(u+1)+ C
Substitute u=e^x
= log(e^x)-log(e^x+1)+ C
= log(e^x/(e^x+1))+ C
= x-log(e^x+1)+ C


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Re: Calc 2 Integral Problem. - February 16th 2012, 06:21 PM

Okay, I got something REALLY close to that but I may have simplified it a little differently/not enough. It was the hardest problem on our test today. YAY! Thank you!


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Re: Calc 2 Integral Problem. - February 16th 2012, 10:52 PM

No problem, PM me if you need another help with calculus or and other math class I've been through all those troubles already.


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