Okay, let’s start out by noting that we are working about $x = 0$ and that means we can use the formula for the Taylor Series of the exponential function. For reference purposes this is,

$${{\bf{e}}^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}}$$

Next, let’s strip out the $n = 0$ term from this and then subtract one. Doing this gives,

$${{\bf{e}}^x} - 1 = \left[ {1 + \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n!}}} } \right] - 1 = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n!}}}$$

Of course, in doing the above step all we really managed to do was eliminate the $n = 0$ term from the series. In fact, that was not a bad thing to have happened as well see shortly.

Finally, let’s divide the whole thing by $x$. This gives,

$$\frac{{{{\bf{e}}^x} - 1}}{x} = \frac{1}{x}\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n!}}} = \sum\limits_{n = 1}^\infty {\frac{{{x^{n - 1}}}}{{n!}}}$$

We moved the $x$ that was outside the series into the series. This is required in order to do the integral of the series. We only want a single $x$ in the problem and we now have that.

Also note that while the function on the left has a division by zero issue the series on the right does not have this problem. All of the $x$’s in the series have positive or zero exponents! This is a really good thing.

Of course, the other good thing that we have at this point is that we’ve managed to find a series representation for the integrand!

Show Step 3

All we need to do now is compute the integral of the series to get a series representation of the integral.

$$\int{{\frac{{{{\bf{e}}^x} - 1}}{x}\,dx}} = \int{{\sum\limits_{n = 1}^\infty {\frac{{{x^{n - 1}}}}{{n!}}} \,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{c + \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{\left( n \right)\left( {n!} \right)}}} }}$$