evaluation of this exponential integral? 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.
