Answer
$F\left( x \right) = \mathop \smallint \limits_0^x \dfrac{{{{\rm{e}}^t} - 1}}{t}{\rm{d}}t = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{x^n}}}{{n\cdot n!}}$
Work Step by Step
From Table 2 of Section 11.7, we have
${{\rm{e}}^t} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{t^n}}}{{n!}} = 1 + t + \dfrac{{{t^2}}}{{2!}} + \dfrac{{{t^3}}}{{3!}} + \dfrac{{{t^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $t$.
So,
${{\rm{e}}^t} - 1 = t + \dfrac{{{t^2}}}{{2!}} + \dfrac{{{t^3}}}{{3!}} + \dfrac{{{t^4}}}{{4!}} + \cdot\cdot\cdot$
$\dfrac{{{{\rm{e}}^t} - 1}}{t} = 1 + \dfrac{t}{{2!}} + \dfrac{{{t^2}}}{{3!}} + \dfrac{{{t^3}}}{{4!}} + \cdot\cdot\cdot$
Integrating both sides:
$\mathop \smallint \limits_0^x \dfrac{{{{\rm{e}}^t} - 1}}{t}{\rm{d}}t = \mathop \smallint \limits_0^x \left( {1 + \dfrac{t}{{2!}} + \dfrac{{{t^2}}}{{3!}} + \dfrac{{{t^3}}}{{4!}} + \cdot\cdot\cdot} \right){\rm{d}}t$
$\mathop \smallint \limits_0^x \dfrac{{{{\rm{e}}^t} - 1}}{t}{\rm{d}}t = \left( {t + \dfrac{{{t^2}}}{{2\cdot 2!}} + \dfrac{{{t^3}}}{{3\cdot 3!}} + \dfrac{{{t^4}}}{{4\cdot 4!}} + \cdot\cdot\cdot} \right)|_0^x$
$\mathop \smallint \limits_0^x \dfrac{{{{\rm{e}}^t} - 1}}{t}{\rm{d}}t = x + \dfrac{{{x^2}}}{{2\cdot 2!}} + \dfrac{{{x^3}}}{{3\cdot 3!}} + \dfrac{{{x^4}}}{{4\cdot 4!}} + \cdot\cdot\cdot$
Thus, $F\left( x \right) = \mathop \smallint \limits_0^x \dfrac{{{{\rm{e}}^t} - 1}}{t}{\rm{d}}t = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{x^n}}}{{n\cdot n!}}$.
