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## #1 2013-02-17 10:11:45

White_Owl
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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 (2013-02-18 11:41:43)

## #2 2013-02-17 10:13:08

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.

## #3 2013-02-17 10:18:34

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.

## #4 2013-02-17 21:26:10

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 2013-02-18 01:46:37

White_Owl
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### Re: integrate (1+2e^x-e^(-x)^(-1)

#### zetafunc. wrote:

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?

zetafunc.
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## #7 2013-02-18 11:12:16

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?

## #8 2013-02-18 11:15:20

anonimnystefy
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### Re: integrate (1+2e^x-e^(-x)^(-1)

You could complete the square in the denominator.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #9 2013-02-18 11:38:02

White_Owl
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### Re: integrate (1+2e^x-e^(-x)^(-1)

mmmm..... Still do not understand.
How do complete the square?

## #10 2013-02-18 12:21:35

anonimnystefy
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### Re: integrate (1+2e^x-e^(-x)^(-1)

You get it to the form (a*x+b)^2+c.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #11 2013-02-19 10:22:44

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.

## #12 2013-02-19 10:38:34

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 (2013-02-19 10:43:46)

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment