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Compute I = $(\int_e^{e^4}\sqrt{\log(x)}dx)$ if it is given that $(\int _1^2 e^{t^2} dt = \alpha )$

I = $([x \sqrt{\log(x)}]_e^{e^4} - \int_e^{e^4} x \frac{1}{2 \sqrt{log(x)}} \frac {1}{x} dx )$
= $([e^4 \sqrt {\log_e e^4} - e \sqrt {\log _e e}] - \frac{1}{2} \int_e^{e^4}\frac{1}{\sqrt{log(x)}} dx )$
= $(2 e^4 - e - \frac{1}{2} \int_e^{e^4}\frac{1}{\sqrt{\log(x)}} dx )$

let $log(x) = (t^2)$

x =$(e^{t^2})$

dx = 2t $(e^{t^2})$ dt

Thus I = $(2 e^4 - e - \frac{1}{2} \int_e^{e^4}\frac{1}{\sqrt{\log(x)}} dx )$

= $(2 e^4 - e - \frac{1}{2} \int _1^2 \frac {1}{t} 2 t e^{t^2} dt )$
= $(2 e^4 - e - \int _1^2 e^{t^2} dt )$
= $(2 e^4 - e - \alpha )$