Answer
$\mathop \smallint \limits_0^x \ln \left( {1 + {t^2}} \right){\rm{d}}t = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{x^{2n + 1}}}}{{n\left( {2n + 1} \right)}}$
Work Step by Step
From Table 2:
$\ln \left( {1 + t} \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{t^n}}}{{n!}} = t - \dfrac{{{t^2}}}{2} + \dfrac{{{t^3}}}{3} - \dfrac{{{t^4}}}{4} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| t \right| \lt 1$ and $t=1$.
Substituting ${t^2}$ for $t$ in the series above gives
$\ln \left( {1 + {t^2}} \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{t^{2n}}}}{n} = {t^2} - \dfrac{{{t^4}}}{2} + \dfrac{{{t^6}}}{3} - \dfrac{{{t^8}}}{4} + \cdot\cdot\cdot$
Taking the definite integral on the series gives
$\mathop \smallint \limits_0^x \ln \left( {1 + {t^2}} \right){\rm{d}}t = \mathop \sum \limits_{n = 1}^\infty \mathop \smallint \limits_0^x \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{t^{2n}}}}{n}{\rm{d}}t = \mathop \sum \limits_{n = 1}^\infty \left[ {\dfrac{{{{\left( { - 1} \right)}^{n - 1}}{t^{2n + 1}}}}{{n\left( {2n + 1} \right)}}} \right]_0^x$
So, $\mathop \smallint \limits_0^x \ln \left( {1 + {t^2}} \right){\rm{d}}t = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{x^{2n + 1}}}}{{n\left( {2n + 1} \right)}}$.
