lol math competition

Okay, can't do these two.

Prove sum(k=1,n){arctan(1/(1+k+k^2))} = arctan(n/(n+2))

Show integral(0,infinity) e^{(x^2 + x^-2)/4} = sqrt(pi/e)

I can't figure out which competition this is from, but this sounds unethical.

1st one sounds like some sort of induction proof.
2nd one sounds like you need to pull some sort of Gauss h4x with a 2D integral.

2nd one sounds incorrect. x^2 + x^-2 > x^2 for all real non-zero x, and integral(0,infinity) e^(x^2/4) = +inf. Ergo integral(0,infinity) e^((x^2 + x^-2)/4) = +inf.

For the first, use arctan(a) + arctan(b) = arctan((a+b)/(1-ab)).

>>3
My 89 says that it's ∞, too. I'd normally trust Mathematica over my calculator, but Mathematica's being a dick tonight—it's spitting out some bullshit long answer, and refuses to simplify or even approximate it. And hell if I'm going to integrate this late at night.

So, are you sure it's √π/e?

e^{(x^2 + x^-2)/4} -> e^{-(x^2 + x^-2)/4}
Mathematica yields right answer.

>>4
Does bullshit long answer involve erfs? lol

>>5
Yes. Yes, it does. And I'm too tired to dick around with it. :-/

>>6
Yeah, you can see the sqrt(pi/e) in the bullshit long expression mathematica spits out. I don't know whether it simplifies over the given bounds.

for the second

it equals e^-(1/2)* integral(0,inf) e^((x+x^-1)/2)^2

so need to show integral e^((x+x^-1)/2)^2=sqrt(pi)

Which is could very well be.

All I could be arsed to do,