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

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