integration question

**robertsays** · April 5th 2008, 11:05 PM

how do you integrate y= e^(-3x)Ln(x)

ive tried integration by parts, but i cant find a way that works.

I'm sure theres a very simple way and i just cant see it.

**deragon999** · April 6th 2008, 12:57 AM

to integrate e divide by the derivative of the power, leave the power the same..

the power is (-3x)Ln(x) which is differentiated using the product rule f'g+g'f
f=-3x f'=-3 g=ln(x) g'=1/x the differential of the power is -3ln(x)+-3x/x which simplifies to -3ln(x)-3

The integral of y= e^(-3x)Ln(x) is therfore y= (e^(-3x)Ln(x))/(-3ln(x)-3)+c

**janvdl** · April 6th 2008, 01:10 AM

Originally Posted by robertsays

how do you integrate y= e^(-3x)Ln(x)

ive tried integration by parts, but i cant find a way that works.

I'm sure theres a very simple way and i just cant see it.

Just to clarify things, do you mean:

$\int e^{-3x \ln x}$

or

$\int e^{-3x} \ln x$

EDIT: Mathematica Integrator says the first one most probably doesn't exist... (When first editing my post I accidentally said the second one doesn't exist. My apologies.)

**deragon999** · April 6th 2008, 02:18 AM

I haven't learnt any way to integrate products other than ones such as trigonometric products where there is a way to get a sum...

Also he hasn't put a x or * to indicate that its e(-3x) times ln(x) so i htink its safe to assume he/she means e(-3x*ln(x))

**janvdl** · April 6th 2008, 02:27 AM

Originally Posted by deragon999

I haven't learnt any way to integrate products other than ones such as trigonometric products where there is a way to get a sum...

Also he hasn't put a x or * to indicate that its e(-3x) times ln(x) so i htink its safe to assume he/she means e(-3x*ln(x))

Mathematica's Integrator finds an answer for the 2nd one... But I cannot find it myself. I tried partial integration.

**Jhevon** · April 6th 2008, 02:56 AM

Originally Posted by deragon999

to integrate e divide by the derivative of the power, leave the power the same..

the power is (-3x)Ln(x) which is differentiated using the product rule f'g+g'f
f=-3x f'=-3 g=ln(x) g'=1/x the differential of the power is -3ln(x)+-3x/x which simplifies to -3ln(x)-3

The integral of y= e^(-3x)Ln(x) is therfore y= (e^(-3x)Ln(x))/(-3ln(x)-3)+c

your method does not work in general and is therefore not correct. in particular, it is not correct in this case (just check by differentiating your answer).

either interpretation yields a nasty integral in this case. one interpretation yields a solution in non-elementary terms, the other can't even be found by the integrator, so it's probably not integrable