# Thread: integrate integral 1/x(ln (x)^1/2)

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

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

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

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

giving integral as

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

when I integrate this last integral I get this

$
\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
$
\int\limits_e^{e^4 } {\frac{1}{{x\sqrt {\ln x} }}} dx
$

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

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

giving integral as

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

when I integrate this last integral I get this

$
\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:

$1^{\frac12} = \dfrac12$

3. doh. lol Silly arithmetic mistakes

Thanks