Compute I = $latex (\int_e^{e^4}\sqrt{\log(x)}dx)$ if it is given that $latex (\int _1^2 e^{t^2} dt = \alpha )$
I = $latex ([x \sqrt{\log(x)}]_e^{e^4} - \int_e^{e^4} x \frac{1}{2 \sqrt{log(x)}} \frac {1}{x} dx )$
= $latex ([e^4 \sqrt {\log_e e^4} - e \sqrt {\log _e e}] - \frac{1}{2} \int_e^{e^4}\frac{1}{\sqrt{log(x)}} dx )$
= $latex (2 e^4 - e - \frac{1}{2} \int_e^{e^4}\frac{1}{\sqrt{\log(x)}} dx )$
let $latex log(x) = (t^2)$
x =$latex (e^{t^2})$
dx = 2t $latex (e^{t^2})$ dt
Thus I = $latex (2 e^4 - e - \frac{1}{2} \int_e^{e^4}\frac{1}{\sqrt{\log(x)}} dx )$
= $latex (2 e^4 - e - \frac{1}{2} \int _1^2 \frac {1}{t} 2 t e^{t^2} dt )$
= $latex (2 e^4 - e - \int _1^2 e^{t^2} dt )$
= $latex (2 e^4 - e - \alpha )$
