1. ## Integral problem

d(x) = *IntegralSign(b^(-x/c) / (f + ax))dx

a, b, c, f are constants

I'm stumped. A little help

2. Mmm... I dunno, I think it doesn't have an elementary primitive.

3. I concur with K. It appears it's not doable by elementary means.

4. I was thinking that if you use u = f+ax and dx = du/a

you might find the solution. I end up with this answer after my math:

ln |ax + f| + C
---------------
ab^(x/c)

I'm not sure if its correct.

5. Originally Posted by cetaamoxclob
d(x) = *IntegralSign(b^(-x/c) / (f + ax))dx

a, b, c, f are constants

I'm stumped. A little help
Originally Posted by cetaamoxclob
I was thinking that if you use u = f+ax and dx = du/a

you might find the solution. I end up with this answer after my math:

ln |ax + f| + C
---------------
ab^(x/c)

I'm not sure if its correct.
The Mathematic site gives the answer in terms of the error function, which can only be approximated. But let's see what your substitution does.

$\displaystyle \int \frac{b^{-x/c}}{f + ax}~dx$

Let $\displaystyle u = f + ax \implies dx = \frac{1}{a}~du$

So
$\displaystyle \int \frac{b^{-x/c}}{f + ax}~dx = \int \frac{b^{-(f - u)/(ac)}}{u}~\frac{du}{a}$

$\displaystyle = \int \frac{b^{-f/(ac)}b^{u/(ac)}}{u}$
(which is pretty much the same form as you started with.)

I can see no way to integrate this (barring some kind of approximation.)

-Dan

6. A friend of mine thinks the answer has something to do with some sort of infinite series.