Answer
$\mathop \smallint \limits_0^x \dfrac{{t - \sin t}}{t}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 3}}}}{{\left( {2n + 3} \right)\left( {2n + 3} \right)!}}$
Work Step by Step
Write $\dfrac{{t - \sin t}}{t} = \dfrac{1}{t}\left( {t - \sin t} \right)$.
From Table 2:
$\sin t = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 1}}}}{{\left( {2n + 1} \right)!}} = t - \dfrac{{{t^3}}}{{3!}} + \dfrac{{{t^5}}}{{5!}} - \dfrac{{{t^7}}}{{7!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $t$.
Thus,
$\dfrac{{t - \sin t}}{t} = \dfrac{1}{t}\left( {t - \sin t} \right) = \dfrac{1}{t}\left( {t - \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} \right)$, ${\ \ \ }$ converges for all $t$.
$\dfrac{{t - \sin t}}{t} = \dfrac{1}{t}\left( {\dfrac{{{t^3}}}{{3!}} - \dfrac{{{t^5}}}{{5!}} + \dfrac{{{t^7}}}{{7!}} + \cdot\cdot\cdot} \right)$
$\dfrac{{t - \sin t}}{t} = \dfrac{{{t^2}}}{{3!}} - \dfrac{{{t^4}}}{{5!}} + \dfrac{{{t^6}}}{{7!}} + \cdot\cdot\cdot = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 2}}}}{{\left( {2n + 3} \right)!}}$
Taking the definite integral on the series gives
$\mathop \smallint \limits_0^x \dfrac{{t - \sin t}}{t}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \mathop \smallint \limits_0^x \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 2}}}}{{\left( {2n + 3} \right)!}}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \left[ {\dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 3}}}}{{\left( {2n + 3} \right)\left( {2n + 3} \right)!}}} \right]_0^x$
So, $\mathop \smallint \limits_0^x \dfrac{{t - \sin t}}{t}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 3}}}}{{\left( {2n + 3} \right)\left( {2n + 3} \right)!}}$.
