Who's up to the challenge? 50

Consider the integral : $\displaystyle J(a)=\int_{0}^{\infty} e^{-(x^2+1)}x^a dx$ , where we want to evaluate $J'(0)$

$\displaystyle J(a) = \frac{\Gamma(\frac{a+1}{2})}{2e}$ , Differentiating with respect to $a$ we get ,

$\displaystyle J'(a) = \frac{1}{4e} [\Gamma(\frac{a+1}{2})\psi_{(0)}(\frac{a+1}{2})]$ , which implies $\displaystyle J'(0)=-\frac{\sqrt{\pi}(\gamma+ln4)}{4e}$

We have , $\displaystyle \int _{ 0 }^{ \infty }{ \frac { \ln { \sqrt [ 50 ]{ x } } }{ { e }^{ { x }^{ 2 }+1 } } dx } = \frac{1}{50}J'(0)=-\frac{\sqrt{\pi}(\gamma+ln4)}{200e}$

Thus making the answer : $\boxed{205}$