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#1 2013-02-16 11:11:45
- White_Owl
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- Registered: 2010-03-03
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integrate (1+2e^x-e^(-x)^(-1)
I have:
using
And now I am stuck.
update: fixed plus signs in the equations
Last edited by White_Owl (2013-02-17 12:41:43)
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#2 2013-02-16 11:13:08
- White_Owl
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Re: integrate (1+2e^x-e^(-x)^(-1)
umm... by some mysterious reason, all plus signs disappeared from under the math tag.
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#3 2013-02-16 11:18:34
- White_Owl
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Re: integrate (1+2e^x-e^(-x)^(-1)
sorry, never mind, I got it.
"Partial fractions" approach is the key.
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#4 2013-02-16 22:26:10
- zetafunc.
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Re: integrate (1+2e^x-e^(-x)^(-1)
Your substitution will work just fine if you use the fact that the integral of 1/(a^2 + x^2) = (1/a)arctan(x/a).
#5 2013-02-17 02:46:37
- White_Owl
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Re: integrate (1+2e^x-e^(-x)^(-1)
Your substitution will work just fine if you use the fact that the integral of 1/(a^2 + x^2) = (1/a)arctan(x/a).
Huh? How does that help here?
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#6 2013-02-17 09:02:13
- zetafunc.
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Re: integrate (1+2e^x-e^(-x)^(-1)
You have a quadratic form in your denominator, what can you do about that?
#7 2013-02-17 12:12:16
- White_Owl
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Re: integrate (1+2e^x-e^(-x)^(-1)
In the denominator I have 2u^2+u-1. The trigonometric substitution requires to have just two elements in the polynom - a squared variable and a squared constant.
Here I have a third member - u, where do you propose it should go?
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#8 2013-02-17 12:15:20
- anonimnystefy
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Re: integrate (1+2e^x-e^(-x)^(-1)
You could complete the square in the denominator.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
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#9 2013-02-17 12:38:02
- White_Owl
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Re: integrate (1+2e^x-e^(-x)^(-1)
mmmm..... Still do not understand.
How do complete the square?
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#10 2013-02-17 13:21:35
- anonimnystefy
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Re: integrate (1+2e^x-e^(-x)^(-1)
You get it to the form (a*x+b)^2+c.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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#11 2013-02-18 11:22:44
- White_Owl
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Re: integrate (1+2e^x-e^(-x)^(-1)
mmmm.....
2u^2+u-1 does not have an (a*x+b)^2+c form or I cannot find it.
2u^2+u-1 = (2u-1)(u+1), and this I used in a "partial fractions" approach. But I cannot find any square form for this particular polynomial.
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#12 2013-02-18 11:38:34
- anonimnystefy
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Re: integrate (1+2e^x-e^(-x)^(-1)
Of course, I am just showing that it can also be done as zf. suggested. The partial fractions work better in this integral.
Last edited by anonimnystefy (2013-02-18 11:43:46)
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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