How to integrate $(e^x-1)/x$

Question

How to integrate $\frac{(e^x-1)}{x}$ and find constant $c$ when $F(0)=0$. I have already tried wolframalpha, but this didn't help to solve this.

Answer 1

Go for infinite series.

$$e^x -1=x+ x^2/2+x^3/6+....$$

$$\frac {e^x-1}{x}=1+x/2+x^2/6+...$$

Integrate and you get $$ \int 1+x/2+x^2/6+....=x+x^2/4+x^3/{18}+....$$

Answer 2

It actually has no closed form integral, but can be expressed using Taylor series as following:$${e^x-1\over x}=\sum_{n=1}^{\infty}{x^{n-1}\over n!}$$Therefore$$\int {e^x-1\over x}dx=\sum_{n=1}^{\infty}{x^{n}\over n\times n!}+C$$

Answer 3

Except for the sign of the argument, this integral is the Ein function, which (as others have said) is not elementary.