Question: > 
> $$\int\frac{e^x+1}{e^x-1}\,\mathrm dx$$
> 
> 
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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.

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

Both terms are of the form $\frac{f'}f$ and you can integrate straight away, giving

$$\log|(e^x-1)(1-e^{-x})|.$$

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Alternatively,

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

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It is questionable whether this is truly "without substitution", because implicitly you are doing

$$\int\frac{f'}{f}dx=\int\frac{df}f.$$
$$\int \frac{e^x}{e^x-1}dx = \int d(\log(e^x-1)) = \log|e^x-1|+C$$
$$\int\frac{e^x}{e^x-1}dx -\int \frac1{e^x-1}dx=\int 1dx = x + C$$
Thus $$\int \frac1{e^x-1}dx=\log|e^x-1|-x+C$$
and finally
$$ \int \frac{e^x+1}{e^x-1}dx = \int \frac{e^x}{e^x-1}dx + \int \frac1{e^x-1}dx= 2\log|e^x-1|-x+C $$