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

Question

$$\int\frac{e^x+1}{e^x-1}\,\mathrm dx$$

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.

Answer 1

Let's use your approach:

$I=\int\frac{e^{x/2}+e^{-x/2}}{e^{x/2}-e^{-x/2}}\mathrm dx=2 \int\frac{\frac{e^{x/2}}{2}+\frac{e^{-x/2}}{2}}{e^{x/2}-e^{-x/2}}\mathrm dx =2 \int\frac{(e^{x/2}-e^{-x/2})'}{e^{x/2}-e^{-x/2}}\mathrm dx =2\ln|e^{x/2}-e^{-x/2}|+c$

Answer 2

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

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

$$\int\frac{f'}{f}dx=\int\frac{df}f.$$

Answer 3

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

Answer 4

$$\int \frac{e^x+1}{e^x-1} dx = \int 1 + \frac 2 {e^x-1} dx = x +2\int \bigg( \frac {e^x}{e^x-1} -1 \bigg)dx$$ Now you apply your formula to evaluate last integral. You get:$$x + 2(\ln|e^x-1| - x) + c$$

Answer 5

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

$$K\int\frac{f'(x)}{f(x)}\,dx=K\ln|f(x)|+C.$$