1. integrate integral 1/x(ln (x)^1/2)

Question asks to evaluate the integral
$\displaystyle \int\limits_e^{e^4 } {\frac{1}{{x\sqrt {\ln x} }}} dx$

Attempt by substitution, Letting u = ln x
$\displaystyle = \int\limits_e^{e^4 } {\frac{1}{x}u^{\frac{{ - 1}}{2}} dx}$

and so
$\displaystyle \begin{array}{l} \frac{{du}}{{dx}} = \frac{1}{x} \\ du = \frac{1}{x}dx \\ \end{array}$

giving integral as

$\displaystyle = \int\limits_e^{e^4 } {u^{\frac{{ - 1}}{2}} du}$

when I integrate this last integral I get this

$\displaystyle \begin{array}{l} = \left[ {\frac{{u^{\frac{1}{2}} }}{{\frac{1}{2}}}} \right]_e^{e^4 } \\ = \left[ {2u^{\frac{1}{2}} } \right]_e^{e^4 } \\ = \left[ {2(\ln x)^{\frac{1}{2}} } \right]_e^{e^4 } \\ = 2(\ln e^4 )^{\frac{1}{2}} - 2(\ln e)^{\frac{1}{2}} \\ = 4 - 1 \\ = 3 \\ \end{array}$

The answer is given as 2, I'm not sure what I have done wrong here ?

2. Originally Posted by Craka
Question asks to evaluate the integral
$\displaystyle \int\limits_e^{e^4 } {\frac{1}{{x\sqrt {\ln x} }}} dx$

Attempt by substitution, Letting u = ln x
$\displaystyle = \int\limits_e^{e^4 } {\frac{1}{x}u^{\frac{{ - 1}}{2}} dx}$

and so
$\displaystyle \begin{array}{l} \frac{{du}}{{dx}} = \frac{1}{x} \\ du = \frac{1}{x}dx \\ \end{array}$

giving integral as

$\displaystyle = \int\limits_e^{e^4 } {u^{\frac{{ - 1}}{2}} du}$

when I integrate this last integral I get this

$\displaystyle \begin{array}{l} = \left[ {\frac{{u^{\frac{1}{2}} }}{{\frac{1}{2}}}} \right]_e^{e^4 } \\ = \left[ {2u^{\frac{1}{2}} } \right]_e^{e^4 } \\ = \left[ {2(\ln x)^{\frac{1}{2}} } \right]_e^{e^4 } \\ = 2(\ln e^4 )^{\frac{1}{2}} - 2\bold{\color{red}{(\ln e)^{\frac{1}{2}}} } \\ = 4 - 1 \\ = 3 \\ \end{array}$

The answer is given as 2, I'm not sure what I have done wrong here ?
You used the somehow rare property:

$\displaystyle 1^{\frac12} = \dfrac12$

3. doh. lol Silly arithmetic mistakes

Thanks