# Thread: Absolute integral

1. ## Absolute integral

Calculate this integral :
They ask for the graphic expliquation .

2. Hi!

Originally Posted by dhiab
Calculate this integral :
use integration by parts

it is $\int u*v' dx = u*v - \int u' v dx$

Let $v':= |ln(x)| => v = \frac{1}{x}$

$u = \frac{1}{x} => u' = - \frac{1}{x^2}$

Thus

$\int \frac{|ln(x)|}{x} dx = \int |ln(x)|*\frac{1}{x} dx$

$= \left[ \displaystyle \frac{1}{x}*\frac{1}{x}\right]^{\displaystyle a}_{\displaystyle 1/a} - \int (-\frac{1}{x^2}*\frac{1}{x}) dx$

You should be able to solve this...

Originally Posted by dhiab
They ask for the graphic expliquation .
What is that supposed to mean?

Rapha

3. Originally Posted by Rapha
Hi!

use integration by parts

it is $\int u*v' dx = u*v - \int u' v dx$

Let $v':= |ln(x)| => v = \frac{1}{x}$

$u = \frac{1}{x} => u' = - \frac{1}{x^2}$

Thus

$\int \frac{|ln(x)|}{x} dx = \int |ln(x)|*\frac{1}{x} dx$

$= \left[ \displaystyle \frac{1}{x}*\frac{1}{x}\right]^{\displaystyle a}_{\displaystyle 1/a} - \int (-\frac{1}{x^2}*\frac{1}{x}) dx$

You should be able to solve this...

Rapha
Hello :
the value of the integral varies according to the absoulue value
What is that supposed to mean? : problem of area

4. Originally Posted by dhiab
Calculate this integral :
They ask for the graphic expliquation .
$\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
$

5. Originally Posted by Rapha
Hi!

use integration by parts

it is $\int u*v' dx = u*v - \int u' v dx$

Let $v':= |ln(x)| => v = \frac{1}{x}$

$u = \frac{1}{x} => u' = - \frac{1}{x^2}$

Thus

$\int \frac{|ln(x)|}{x} dx = \int |ln(x)|*\frac{1}{x} dx$

$= \left[ \displaystyle \frac{1}{x}*\frac{1}{x}\right]^{\displaystyle a}_{\displaystyle 1/a} - \int (-\frac{1}{x^2}*\frac{1}{x}) dx$

You should be able to solve this...
HELLO :

6. Originally Posted by dhiab
HELLO :
yes, this is correct. it is why Plato's answer has two integrals in it. as his post indicates, a good old fashioned u-substitution of $u = \ln x$ gets the job done. make sure you understand what he did and finish up the problem