# Math Help - evaluate integral

1. ## evaluate integral

hello,im kinda stuck with this integral,
i wrote my approach,but it didnt get me too far.
thanks!

2. Ermmm...

3. i suppose there is no analytic solution for that thing then?
how much reliable really is that integrator?

4. Originally Posted by becker89
i suppose there is no analytic solution for that thing then?
how much reliable really is that integrator?
It's Wolfram - possibly the best 'automatic' integrator in the world. I just tested it in Matlab, and the answer given there is equally as dubious. In fact Matlab says "warning, explicit integral could not be found." So I'd definitely say, yes, there is no analytic solution.

5. Where did you find such a question anyway?

6. well i have another integral,i think putting it here is better than opening a new thread.
this integral is something unique,according to the person who uploaded it,
there is a solution using elementary function,
even though the fact that if you plug it in to different integrators,
it would show a messy result (like the integral i asked about earlier).

the factoring i did is a suggestion i got from the same person
many thanks!

7. Originally Posted by Mush
Where did you find such a question anyway?
someone uploaded it on another forum im in

8. The four complex roots of t^4 - 2t^3 + t^2 + 6t + 3 are

[sqrt(6) + 1 ]/2 +/- [sqrt(3) + sqrt(2) ]/2 x sqrt(-1) and
[1 - sqrt(6)]/2 +/- [ sqrt(3) - sqrt(2) ] /2 x sqrt(-1)

They are two conjugate pairs so that
t^4 - 2t^3 + t^2 + 6t + 3 can be factorized into two quadartic polynomials .

.........

the result i found is

(k1) ln| t^2 - [sqrt(6) + 1]t + 3 + sqrt(6) | +
(k2) ln| t^2 - [1 - sqrt(6)]t + 3 - sqrt(6) | +
(k3) arctan{ [sqrt(3) - sqrt(2)][2t - sqrt(6) - 1 ] } +
(k4) arctan{ [sqrt(3) + sqrt(2)][2t + sqrt(6) - 1] } + C

where k1,k2,k3,k4 are constant and t = x + sqrt ( x^2 + x + 1)
I'm not sure ...
k1 = -0.237
k2 = -0.0160
k3 = -0.823
k4 = -0.0555