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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