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.