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

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

umm... by some mysterious reason, all plus signs disappeared from under the math tag.

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

sorry, never mind, I got it.

"Partial fractions" approach is the key.

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**zetafunc.****Guest**

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).

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

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.****Guest**

You have a quadratic form in your denominator, what can you do about that?

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

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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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,981

You could complete the square in the denominator.

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

mmmm..... Still do not understand.

How do complete the square?

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,981

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

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

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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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,981

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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