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.
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
Just to clarify things, do you mean:
$\displaystyle \int e^{-3x \ln x} $
or
$\displaystyle \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.)
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))
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
angel.white, try this integrator too: The Integrator--Integrals from Mathematica