Why can't I compute the integral with the following method?

Question

I understand the following solution, where we multiply the integral by $2/2$ so we can make use of the standard form to solve it. $$\int \frac{e^x}{2e^x-1} \, \mathrm{d}x = \frac{1}2 \int \frac{2e^x}{2e^x-1} \, dx = \frac {1}2 \ln|2e^x-1|+C$$ However, I tried doing this integral with another method where I took out the constant and used the standard form to integrate, but it does not work. Why? $$\int \frac{e^x}{2e^x-1} \, \mathrm{d}x = \int \frac{e^x}{2(e^x-\frac{1}2)} \, dx = \frac {1}2 \int \frac{e^x}{e^x-\frac{1}2} = \frac {1}2 \ln \left|e^x-\frac{1}2 \right| +C$$

Answer 1

Your answers are equivalent:

$$\frac 1 2 \ln |2 e^x - 1| = \frac 1 2 \left( \ln(2) + \ln \left| e^x - \frac 12 \right| \right)$$

The two indefinite integrals differ by a constant (namely, $\ln(2)/2$), and hence everything is fine. You may think of it as being the case that

$$ \begin{align*} &\int \frac{e^x}{2e^x-1} \, \mathrm{d}x = \frac {1}2 \ln|2e^x-1|+C \\ &\int \frac{e^x}{2e^x-1} \, \mathrm{d}x = \frac {1}2 \ln \left|e^x-\frac{1}2 \right| +D \end{align*}$$

where $C,D$ are constants of integration, and $D = C + \ln(2)/2$ (still a constant, but just a shifted one).

You can verify this by differentiation as well.

Bear in mind what's going on here: any two antiderivatives of a function differ by a constant, as a consequence of the mean value theorem. Thus, when you write "$F(x)+C$" for an integral, you're not saying that any one function $F$ is an antiderivative, but rather any function that differs from $F$ by a constant is an antiderivative.