Question

5Cleft%28%20%20%20%5Cint%5E%7B1%7D_0%20%7Be%7D%5E%7B%20%5Csqrt%5B%5D%7B%20%7Be%7D%5E%7Bx%7D%20%7D%20%7D%20%20%20%5C%3A%20dx%20%2B%202%20%20%20%5Cint_%7Be%7D%5E%7B%20%7Be%7D%5E%7B%20%5Csqrt%7Be%7D%20%7D%20%7D%20ln%28%20ln%28x%29%20%29%20%20%20%5C%3A%20dx%5Cright%29%20%20%5C%5C%20" id="TexFormula1" title=" \rm \frac{ {10}^5 }{ {e}^{ \sqrt{e} } } \left( \int^{1}_0 {e}^{ \sqrt[]{ {e}^{x} } } \: dx + 2 \int_{e}^{ {e}^{ \sqrt{e} } } ln( ln(x) ) \: dx\right) \\ " alt=" \rm \frac{ {10}^5 }{ {e}^{ \sqrt{e} } } \left( \int^{1}_0 {e}^{ \sqrt[]{ {e}^{x} } } \: dx + 2 \int_{e}^{ {e}^{ \sqrt{e} } } ln( ln(x) ) \: dx\right) \\ " align="absmiddle" class="latex-formula">​

Answer 1

In the first integral, substitute $x \to e^{\sqrt{e^x}}$:

$\displaystyle I = \int_0^1 e^{\sqrt{e^x}} \, dx = 2 \int_e^{e^{\sqrt e}} \frac{dx}{\ln(x)}$

In the second integral, integrate by parts:

$\displaystyle J = \int_e^{e^{\sqrt e}} \ln(\ln(x)) \, dx = \dfrac12 e^{\sqrt e} - \int_e^{e^{\sqrt e}} \frac{dx}{\ln(x)}$

It follows that

$\dfrac{10^5}{e^{\sqrt e}}(I+2J) = \dfrac{10^5}{e^{\sqrt e}} \times e^{\sqrt e} = \boxed{10^5}$