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Topic review (newest first)
- anonimnystefy
- 2013-02-19 10:38:34
Of course, I am just showing that it can also be done as zf. suggested. The partial fractions work better in this integral.
- White_Owl
- 2013-02-19 10:22:44
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.
- anonimnystefy
- 2013-02-18 12:21:35
You get it to the form (a*x+b)^2+c.
- White_Owl
- 2013-02-18 11:38:02
mmmm..... Still do not understand.
How do complete the square?
- anonimnystefy
- 2013-02-18 11:15:20
You could complete the square in the denominator.
- White_Owl
- 2013-02-18 11:12:16
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?
- zetafunc.
- 2013-02-18 08:02:13
You have a quadratic form in your denominator, what can you do about that?
- White_Owl
- 2013-02-18 01:46:37
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.
- 2013-02-17 21:26:10
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
- 2013-02-17 10:18:34
sorry, never mind, I got it.
"Partial fractions" approach is the key.
- White_Owl
- 2013-02-17 10:13:08
umm... by some mysterious reason, all plus signs disappeared from under the math tag.
- White_Owl
- 2013-02-17 10:11:45
I have:
using
And now I am stuck.
update: fixed plus signs in the equations