Is this your integral : ?

Results 1 to 6 of 6

- June 6th 2011, 10:35 AM #1

- Joined
- Jun 2011
- Posts
- 11

- June 6th 2011, 11:28 AM #2

- June 6th 2011, 12:09 PM #3

- Joined
- Jun 2011
- Posts
- 11

- June 6th 2011, 12:49 PM #4

- Joined
- Oct 2008
- Posts
- 1,035
- Thanks
- 49

Bit of give and take in the numerator...

Do parts on

(integrating the fraction)...

Edit: Just in case a picture helps...

... where (key in spoiler) ...

__Spoiler__:

Larger

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

- June 6th 2011, 01:42 PM #5

- Joined
- Jun 2011
- From
- Right now it's the Earth, a strange place
- Posts
- 17

What a strange integral... considering the amount of trouble you have to go through to solve it (using any "normal" method) you'd think this can't have a nice solution but it has (I used Mathematica - sorry if that feels for you like I cheated) and wound up with something that was... well, funny how "nice" it was.

Anyway, all you have to do is note that the form tom@ballooncalculus wrote can be rewritten as:

And that is practically the definition of the quotient rule, so:

Of course you get the same result by substituting

.

Now, since any other way I can think of would require evaluating

, I guess this is the way it's supposed to be solved. I also congratulate anyone who would have been able to do that .

- June 6th 2011, 03:37 PM #6

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,444
- Thanks
- 787

one might guess, because the numerator is a square, that the integral may be some quotient.

differentiating gives:

, which we want to be equal to

that is, we want .

if f(x) is a polynomial, with leading term of degree n:

, then the leading term of

is .

this tells us n = 1, and a = -1, that is f(x) = -x + b, for some real number b.

then

, which only holds when b = -1.

we can thus conclude that f(x) = -x - 1