Answer
$$s\left( t \right) = t - {\tan ^{ - 1}}t{\text{ and }}v\left( t \right) = 1 - \frac{1}{{{t^2} + 1}}$$
Work Step by Step
$$\eqalign{
& {\text{Use theorem 6}}{\text{.2 }} \cr
& {\text{Given the acceleration }}a\left( t \right){\text{ of an object moving along a line and its initial velocity}} \cr
& {\text{ }}v\left( 0 \right){\text{, the velocity of the object for future time }}t \geqslant 0{\text{ is }}v\left( t \right) = v\left( 0 \right) + \int_0^t {a\left( x \right)} dx \cr
& {\text{let }}a\left( x \right) = \frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{ and }}v\left( 0 \right) = 0.{\text{ then }}v\left( t \right){\text{ is}} \cr
& v\left( t \right) = 0 + \int_0^t {\frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr
& or \cr
& v\left( t \right) = \int_0^t {{{\left( {{x^2} + 1} \right)}^{ - 2}}\left( {2x} \right)} dx \cr
& {\text{integrate using }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& v\left( t \right) = \left( {\frac{{{{\left( {{x^2} + 1} \right)}^{ - 1}}}}{{ - 1}}} \right)_0^t \cr
& v\left( t \right) = - \left( {\frac{1}{{{x^2} + 1}}} \right)_0^t \cr
& {\text{evaluate the limits}} \cr
& v\left( t \right) = - \left( {\frac{1}{{{t^2} + 1}} - \frac{1}{{{0^2} + 1}}} \right) \cr
& {\text{simplify}} \cr
& v\left( t \right) = 1 - \frac{1}{{{t^2} + 1}} \cr
& \cr
& {\text{Use theorem 6}}{\text{.1 }} \cr
& {\text{Given the velocity }}v\left( t \right){\text{ of an object moving along a line and its initial position}} \cr
& {\text{ }}s\left( 0 \right){\text{, the position function of the object for future time }}t \geqslant 0{\text{ is }}s\left( t \right) = s\left( 0 \right) + \int_0^t {v\left( x \right)} dx \cr
& {\text{let }}v\left( x \right) = 1 - \frac{1}{{{x^2} + 1}}{\text{ and }}s\left( 0 \right) = 0.{\text{ then }}s\left( t \right){\text{ is}} \cr
& s\left( t \right) = 0 + \int_0^t {\left( {1 - \frac{1}{{{x^2} + 1}}} \right)} dx \cr
& {\text{integrate }} \cr
& s\left( t \right) = \left( {x - {{\tan }^{ - 1}}x} \right)_0^t \cr
& {\text{evaluate the limits}} \cr
& s\left( t \right) = \left( {t - {{\tan }^{ - 1}}t} \right) - \left( {0 - {{\tan }^{ - 1}}0} \right) \cr
& {\text{simplify}} \cr
& s\left( t \right) = t - {\tan ^{ - 1}}t \cr} $$
