Infinite dimensions, not quite infinite volume

First, we evaluate the integral using the substitution $u=x+y^2$ : \left(1+6\int_{0}^B \frac{y\left(x+y^2\right)^2}{x^3}dy\right)^x= \left(1+\frac{\left(x+y^2\right)^3}{x^3}\bigg{|}_0^B\right)^x= \left(1+\frac{\left(x+B^2\right)^3}{x^3}-1\right)^x=\left(\frac{\left(x+B^2\right)}{x}\right)^{3x}

This is an indeterminate form of the type $1^\infty$ , so we will convert to to $\exp \left(\ln \left(\frac{\left(x+B^2\right)}{x}\right)^{3x}\right)=\exp \left(3x\ln \left(\frac{\left(x+B^2\right)}{x}\right)\right)=\exp \left(3x\left(\ln \left(x+B^2\right)-\ln x\right)\right)$

The equation inside the $\exp$ is now an indeterminate form of the type $0\cdot\infty$ , so we will modify it to be $\exp \left(\frac{\ln \left(x+B^2\right)-\ln x}{\frac{1}{3x}}\right)$ then apply L'Hospital's Rule:

$\exp \left(\frac{\ln \left(x+B^2\right)-\ln x}{\frac{1}{3x}}\right)=\left(\frac{\frac{1}{x+B^2}-\frac{1}{x}}{-\frac{1}{3x^2}}\right)=\exp \left(-\frac{3x^2}{x+B^2}+3x\right)$

Long dividing the first term, we get

$-3x+3B^2-\frac{3B^4}{x+B^2}+3x=3B^2-\frac{3B^4}{x+B^2}$

Finally, because x is going to $\infty$ , we obtain the limit as $e^{3B^2}$

We are looking for the term in the exponent to be 2016.

$3B^2=2016$

$B^2=672$

$B=\sqrt{672}=4\sqrt{42}$

$\alpha=4, \beta=42, \alpha+\beta=\boxed{46}$