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## #1 2013-02-16 11:11:45

White_Owl
Member
Registered: 2010-03-03
Posts: 106

### 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
Member
Registered: 2010-03-03
Posts: 106

### 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
Member
Registered: 2010-03-03
Posts: 106

### 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.
Guest

### 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
Member
Registered: 2010-03-03
Posts: 106

### 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?

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zetafunc.
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## #7 2013-02-17 12:12:16

White_Owl
Member
Registered: 2010-03-03
Posts: 106

### 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
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### 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
The knowledge of some things as a function of age is a delta function.

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## #9 2013-02-17 12:38:02

White_Owl
Member
Registered: 2010-03-03
Posts: 106

### 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
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### 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
Member
Registered: 2010-03-03
Posts: 106

### 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
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### 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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