The problem is that the logarithm of a negative number is complex. However, the rules of logarithms still apply, so we see that
$$
I = \frac{25000}{\log 2}\Big(\log(-32) + \log(99999+16^x)\Big)
$$
The $\log(-32)$ part, although complex, is a constant of integration which has been chosen by the integrator. Thus, the integral is of the form
$$
I = \frac{25000\log(99999+16^x)}{\log 2} + c
$$
This is perfectly real for real choices of $c$ and direct computation shows that this differentiates down to our starting function.