how could i evaluate
$$ \int_{0}^{a(E)}\sqrt{E-16\pi ^{2}e^{4x}} $$
where 'a' is the point so $ E-16\pi^{2}e^{4a}=0 $
this appear s in Quantum mechanic so i think the answer is something like
$$ E^{1/2}log(E) $$ with some constants but it goes like this
Consider the problem of the antiderivative $$\int\sqrt{E-16\pi ^{2}e^{4x}}\, dx$$ Change variable $$E-16\pi ^{2}e^{4x}=y^2\implies x=\frac{1}{4} \log \left(\frac{E-y^2}{16 \pi ^2}\right)\implies dx=-\frac{y}{2 \left(E-y^2\right)}\,dy$$ All of this makes $$I=-\int \frac{y^2}{2 \left(E-y^2\right)}\,dy=\int \left(\frac{E}{2 \left(y^2-E\right)}+\frac{1}{2}\right)\,dy$$
I am sure that you can take it from here.
For any $K>1$ we have: $$\begin{eqnarray*}\int_{0}^{\log K}\sqrt{K-e^{x}}\,dx =\int_{1}^{K}\frac{\sqrt{K-u}}{u}\,du&=&\sqrt{K}\int_{1/K}^{1}\sqrt{1-u}\,\frac{du}{u}\\&=&2\sqrt{K}\left(\text{arctanh}\left(\sqrt{1-\frac{1}{K}}\right)-\sqrt{1-\frac{1}{K}}\right)\\&=&\sqrt{K}\log\left(\frac{\sqrt{K}+\sqrt{K-1}}{\sqrt{K}-\sqrt{K-1}}\right)-2\sqrt{K-1}\\[0.2cm]&=&\color{red}{2\sqrt{K}\log(\sqrt{K}+\sqrt{K-1})-2\sqrt{K-1}}\end{eqnarray*}$$ hence you are right.
