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

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