I've seen that $I$ satisfies that $$\lim_{x\to 0^{+}}I(x)=0$$and also that$$\lim_{x\to+\infty}I(x)=\lim_{x\to\infty}\frac{1}{\sqrt{e^{x}}}\int_{0}^{x}\frac{dt}{\sqrt{1-\frac{e^{t}}{e^{x}}}}=0,$$ so by Rolle's theorem there must be at least one maximum in $(0,+\infty)$. And I have to prove that there is only one maximum, I just don't seem to be able to do it because when I use Leibniz's integral rule I get that $$I'(x)=\frac{1}{\sqrt{e^{x}-e^{x}}}+\int_{0}^{x}\frac{e^{x}}{2\sqrt{\left(e^{x}-e^{t}\right)^{3}}}dt$$ which explodes because of the first term. I don't know how to get further than this step.
Although it's better to have a mathematical proof, it would also be helpful to be able to plot the function on MatLab, it just keeps giving me dimension errors somehow. All help will be welcome.
$$I=\int \frac{dt}{\sqrt{e^{x}-e^{t}}}=\int \frac{dt}{e^{x/2}\sqrt{1-e^{(t-x)}}}$$ $$t=x+\log(u) \implies dt=\frac {du}u\implies I=e^{x/2}\int\frac {du}{u \sqrt{1-u} }$$ $$u=1-v^2 \implies \int\frac {du}{u \sqrt{1-u} }=2\int \frac {dv}{1-v^2}=-2 \tanh ^{-1}(v) $$
Back to $x$ and using the bounds leads to $$J(x)=\int_0^x \frac{dt}{\sqrt{e^{x}-e^{t}}}=2 e^{-x/2} \tanh ^{-1}\left(e^{-x/2} \sqrt{e^x-1}\right)$$ $$J'(x)=\frac{1}{\sqrt{e^x-1}}-e^{-x/2} \tanh ^{-1}\left(e^{-x/2} \sqrt{e^x-1}\right)$$
By inspection or graphing, the solution is just above $x=1$ and Newton methods gives the folloxing iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.0000000 \\ 1 & 1.1597496 \\ 2 & 1.1862399 \\ 3 & 1.1868419 \\ 4 & 1.1868422 \end{array} \right)$$ and, at this point, $J''(x)=-0.47692$ confirms that this is a maximum.