Fake proof that $\frac{e^x-1}{e^x+1}=e^x$, via integrating $\operatorname{sech} x$ in two ways We start with the integral:
$$\int \text{sech}(x)dx$$  

Method 1
\begin{align}
\int \text{sech}(x)dx & = \int\frac{2}{e^x+e^{-x}}dx \\
&= \int\frac{2e^x}{e^{2x}+1}dx
\end{align}
Using the substitution $u=e^x$,
\begin{align}
\int \text{sech}(x)dx & = \int\frac{2}{u^2+1}du \\
&= 2\text{ arctan}(u)+c \\
&= 2\text{ arctan}(e^x)+c
\end{align}

Method 2
\begin{align}
\int \text{sech}(x)dx & = \int\frac{\text{cosh}(x)}{\text{cosh}^2(x)}dx \\
& = \int\frac{\text{cosh}(x)}{\text{sinh}^2(x)+1}dx
\end{align}
Using the substitution $u=\text{sinh}(x)$,
\begin{align}
\int \text{sech}(x)dx & = \int\frac{1}{u^2+1}du \\
&= \text{arctan}(\text{sinh}(x))+c \\
&= 2\text{ arctan}(\text{tanh}(\frac{x}{2}))+c \\
&= 2\text{ arctan}(\frac{e^x-1}{e^x+1})+c
\end{align}

Thus, we obtain:  

$$\frac{e^x-1}{e^x+1}=e^x$$  

However, I see no reason why they are equal. Did I do something wrong in the calculation?

Because $2\arctan\left(\frac{e^x-1}{e^x+1}\right)=2\left(\arctan(e^x)-\frac\pi4\right)=2\arctan(e^x)+C'$.
The results differ by a constant.
