This is making me extremelly pissed off, because I saw a similliar integral that was apparently unsolvable, and now dear prof send this in the list without any resolution or help.
The whole question is this:
Consider the region S bounded by $y=1+\sqrt{x}e^{x^2}$, $y=1$, $x=a$ (x=a is in the right side of the Y-axis). Calculate the value of "A", so the volume of the solid generated by the rotation of S around $y=1$ be equal to $2\pi$.
Since $V=\int_0^a\pi(\sqrt{x}e^{x^2})^2\;dx=\pi\int_0^a x e^{2x^2}\;dx=\pi[\frac{1}{4}e^{2x^2}]_0^a=\frac{\pi}{4}(e^{2a^2}-1)=2\pi$,
$e^{2a^2}=9$ so $2a^2=\ln9=2\ln3\implies a^2=\ln3\implies a=\sqrt{\ln 3}$.