I am looking for a function f(x) that by itself is continues, and its integral is infinity but when multiply by $1/\log(1+e^{-x})$ makes it integrable. In other words,
$$ \int_R \frac{1}{\log(1+e^{-x})} f(x) dx <\infty $$ where
$$ \int_R f(x) dx =\infty. $$
Basically what integrability is about given that the function is continuous is that it has to approach zero fast enough. In this case I note that for negative $x$ we get:
$1/\log(1+e^{-x})\approx 1/\log(e^{-x}) = {1\over-x}$
This suggest that a function that behaves as $1/x$ for negative $x$ will not be integrable (as you know $1/x$ is not integrable because it doesn't converge fast enough), but on the other hand $1/x^2$ converges fast enough.
So a candidate for $f$ would be for example:
$f(x) = \begin{cases}-1/x & \hbox{ if }& x<-1\\ -x &\hbox{ if } & -1\le x<0 \\ 0& \hbox{otherwise}\end{cases}$
It's quite easy to prove that $\int f(x) dx$ doesn't converge. To prove that $\int f(x) dx/\log(1+e^{-x})$ converges one could use the estimate $0\le f(x)/\log(1+e^{-x}) \le f(x)/\log(e^{-x}) = f(x)/-x$ and use that $-\int f(x)dx/x$ converges (here we use the fact that we constructed $f(x)$ so that the integrand will vanish for non-negative $x$).