Logarithmic part of the Risch Alorithm I'm reading some paper about the Risch algorithm and wanted to try a little example: 
I want to find an elementary solution for:
$$\int\frac{1}{e^x + 1}$$
The following lemma tells me how to do this: 

and the following tells me how to calculate the $c_i$ values:


My fraction $\frac{1}{e^x + 1}$ is a proper fraction, the denominator is square-free and not divisible by $\theta = e^x$ and $\int\frac{1}{e^x + 1}$ is elementary (Wolfram Alpha). 
Using this approach I get $$1 = c_1 \cdot \theta^{'} = c_1 \cdot e^x$$
Therefor $c_1 = 1 / e^x$. But $c_1$ is supposed to be a constant in $\mathbb{Q}$. What am I doing wrong?

The solution to the integration should be
$$\int\frac{1}{e^x + 1} = x - log(e^x + 1)$$
It looks similar to the form of the lemma, but from where comes the $x$?
 A: Or... Ad and subtract an e-power in the numerator, then divide the fraction. You will integrate $1$ and $\frac{-e^{x}}{1+e^x}$ The latter one is an $ln$ term. 
A: 
I want to find an elementary solution for $\displaystyle\int\frac{1}{e^x + 1}$

Normally, I wouldn't be upset at someone for writing $\displaystyle\int f(x)$ instead of $\displaystyle\int f(x)\color{red}{dx}$, but it would appear that in this particular case the lack of a proper $dx=d(\ln\theta)\neq d\theta$ is the source of all your confusion. After all, your integral is not $\displaystyle\int\frac1{1+\theta}d\theta=1\cdot\ln(\theta+1)$, but rather $\displaystyle\int\frac1{1+\theta}d(\ln\theta)$, which is a completely different thing altogether, wouldn't you agree ?

My fraction $\dfrac1{e^x+1}$ is a proper fraction, the denominator is square-free and not divisible by $\theta=e^x$.

If $\theta=e^x$, then your integral becomes $\displaystyle\int\frac1{1+e^x}d(e^x)=\int\frac{e^x}{1+e^x}dx\neq\int\frac1{1+e^x}dx$. Please notice that, were this the case, then the formula would actually apply, leading to $\ln(e^x+1)$.
