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Of course, I am just showing that it can also be done as zf. suggested. The partial fractions work better in this integral.
You get it to the form (a*x+b)^2+c.
mmmm..... Still do not understand.
You could complete the square in the denominator.
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.
You have a quadratic form in your denominator, what can you do about that?
Huh? How does that help here?
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).
sorry, never mind, I got it.
umm... by some mysterious reason, all plus signs disappeared from under the math tag.
And now I am stuck.
update: fixed plus signs in the equations