How to compute $\int\tfrac{e^x+1}{e^x-1}\,\mathrm dx$ without substitution? 
$$\int\frac{e^x+1}{e^x-1}\,\mathrm dx$$

For this above problem, I tried adding and subtracting a $e^x$ to the numerator and proceeded. I did end up with an answer. I also tried to solve this question by taking $e^{x/2}$ common from both numerator and the denominator.
$$\int\frac{e^{x/2}+e^{-x/2}}{e^{x/2}-e^{-x/2}}\mathrm dx$$
Once I've taken $e^{x/2}$ out from the denominator and the numerator, I thought of applying this property:
$$\int\frac{f'(x)}{f(x)}\, \mathrm dx = \ln |f(x)|+C$$
But I am unable to manipulate my obtained expression using this property.
 A: $$\frac{e^x+1}{e^x-1}=\frac{e^x}{e^x-1}+\frac{e^{-x}}{1-e^{-x}}.$$
Both terms are of the form $\frac{f'}f$ and you can integrate straight away, giving
$$\log|(e^x-1)(1-e^{-x})|.$$

Alternatively,
$$\frac{e^x+1}{e^x-1}=\frac{e^{x/2}+e^{-x/2}}{e^{x/2}-e^{-x/2}}=2\frac{(e^{x/2}-e^{-x/2})'}{e^{x/2}-e^{-x/2}}\to 2\log|e^{x/2}-e^{-x/2}|.$$

It is questionable whether this is truly "without substitution", because implicitly you are doing
$$\int\frac{f'}{f}dx=\int\frac{df}f.$$
A: $$\int \frac{e^x+1}{e^x-1} dx = \int 1 + \frac 2 {e^x-1} dx = x +2\int \bigg( \frac {e^x}{e^x-1} -1 \bigg)dx$$
Now you apply your formula to evaluate last integral. You get:$$x + 2(\ln|e^x-1| - x) + c$$
A: $$\int \frac{e^x}{e^x-1}dx = \int d(\log(e^x-1)) = \log|e^x-1|+C$$
$$\int\frac{e^x}{e^x-1}dx -\int \frac1{e^x-1}dx=\int 1dx = x + C$$
Thus $$\int \frac1{e^x-1}dx=\log|e^x-1|-x+C$$
and finally
$$ \int \frac{e^x+1}{e^x-1}dx = \int \frac{e^x}{e^x-1}dx + \int \frac1{e^x-1}dx= 2\log|e^x-1|-x+C $$
A: Multiply the numerator and denominator by $e^{-x/2}$ to form
$$\frac{e^x+1}{e^x-1}\times\frac{e^{-x/2}}{e^{-x/2}}=\frac{e^{x/2}+e^{-x/2}}{e^{x/2}-e^{-x/2}}=\frac{\cosh(\frac{x}{2})}{\sinh(\frac{x}{2})},$$
then
$$\int \frac{e^x+1}{e^x-1}\,dx=\int \frac{\cosh(\frac{x}{2})}{\sinh(\frac{x}{2})}\,dx=2\int\frac{d\left(\sinh\left(\frac{x}{2}\right)\right)}{{\sinh(\frac{x}{2})}}=2\ln\left|\sinh\left(\frac{x}{2}\right)\right|+C.$$
If $K$ is a constant, then all integrals of the form $\frac{f'}{f}$ evaluate to
$$K\int\frac{f'(x)}{f(x)}\,dx=K\ln|f(x)|+C.$$
A: Let's use your approach:
$I=\int\frac{e^{x/2}+e^{-x/2}}{e^{x/2}-e^{-x/2}}\mathrm dx=2 \int\frac{\frac{e^{x/2}}{2}+\frac{e^{-x/2}}{2}}{e^{x/2}-e^{-x/2}}\mathrm dx =2 \int\frac{(e^{x/2}-e^{-x/2})'}{e^{x/2}-e^{-x/2}}\mathrm dx =2\ln|e^{x/2}-e^{-x/2}|+c$
