If the value of $\int_{1}^2{e^{x^{2}}}dx$ is $\alpha$, then what is $$\int_{e}^{e^{4}}{\sqrt{\log x}}dx$$
It just seems that substitution of any sort does not help. Is there some other way in which these two integrals are related?
The answer is $2e^4 - e - \alpha$
Take $t=\sqrt{\log x}$ then $x=e^{t^2}$ and $dx=2te^{t^2}$. So $$I=\int_e^{e^4}\sqrt{\log x}dx=\int_1^22t^2e^{t^2}dt. $$ Using integration by parts: $u=t$ $dv=2te^{t^2}dt$, you'll get that $v=e^{t^2}$ and $$I=te^{t^2}\bigg|_1^2-\int_1^2e^{t^2}dt=2e^4-e-\alpha $$
$$y=e^{x^2}\iff log y =x^2\iff \sqrt{\log y}=x$$ The rectangle bounded by $(1,0), (2,0),(1,e^4), (2,e^4)$ has area $e^4.$ The regions represented by the two integrals fill this rectangle with an excess of a $1\times(e^4-e)$-rectangle that touches the Y-axis. (In the second integral, interchange $x$ and $y$ and use the equivalence displayed above.) Therefore, $$\int_{e}^{e^{4}}{\sqrt{\log x}}\,dx=e^4 + (e^4-e)-a=2e^4 -e-a.$$
