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

**Craka** · October 30th 2008, 03:05 AM

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 ?

**earboth** · October 30th 2008, 03:21 AM

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:

$1^{\frac12} = \dfrac12$

**Craka** · October 30th 2008, 03:25 AM

doh.

lol Silly arithmetic mistakes

Thanks

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

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

Integrate the following, definite integral

How to integrate this integral?

integrate

Integrate a function which includes the integral of itself

integrate

Search Tags