Calculate this integral :
They ask for the graphic expliquation .
Hi!
use integration by parts
it is $\displaystyle \int u*v' dx = u*v - \int u' v dx$
Let $\displaystyle v':= |ln(x)| => v = \frac{1}{x}$
$\displaystyle u = \frac{1}{x} => u' = - \frac{1}{x^2}$
Thus
$\displaystyle \int \frac{|ln(x)|}{x} dx = \int |ln(x)|*\frac{1}{x} dx $
$\displaystyle = \left[ \displaystyle \frac{1}{x}*\frac{1}{x}\right]^{\displaystyle a}_{\displaystyle 1/a} - \int (-\frac{1}{x^2}*\frac{1}{x}) dx$
You should be able to solve this...
What is that supposed to mean?
Rapha
$\displaystyle \int\limits_{1/e}^e {\frac{{\left| {\ln (x)} \right|}}
{x}dx} = \int\limits_{1/e}^1 {\frac{{ - \ln (x)}}
{x}dx} + \int\limits_1^e {\frac{{\ln (x)}}
{x}dx = } \left. {\frac{{ - \ln ^2 (x)}}
{2}} \right|_{1/e}^1 + \left. {\frac{{\ln ^2 (x)}}
{2}} \right|_1^e
$