Recently I asked a question where as answer came out $\ln{(a)}=2*\text{arctanh } u$ which is to be calculated using $\sum_{k=0}^{\infty}(\frac{c^{2k+1}}{2k+1})$ with $u=\frac{a-1}{a+1} < 1$ if that helps.
Now since I need to approach and not necessarily calculate $\ln{a}$, I figured that if I used it like the Riemann-sums, the antiderivative of the function can also be used, since the sum is leading to an area of $1*f(x)$ and the antiderivative will approximately reach that value (not quite exact, but like I said, that was not necessary).
Anyways, my question was, what is the antiderivative of $f(x)=2* \frac{c^{2x+1}}{2x+1}$? I don't really know if it's very hard, I'm just very bad at calculating antiderivatives when they're not as simple :s
I’m afraid the solution isn’t elementary. But,
$$I = \int 2 \frac{c^{2x+1}}{2x+1}\, dx = \int \frac{c^u}{u}\, du$$ substituting $u = 2x+1$. Now, this is an exponential integral given by: $$I = \operatorname{Ei}\bigl((2x+1)\ln c\bigr) + C$$