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A nice problem from ISI 10+2

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 )

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