Answer
$s(t) = t^2-tan^{-1}~t+1$
Work Step by Step
$s(t)$ is the position of the particle at time $t$
$\frac{ds}{dt} = v(t)$ is the velocity of the particle at time $t$
$v(t) = 2t-\frac{1}{(1+t^2)}$
$s(t) = \int [2t-\frac{1}{(1+t^2)}]~dt = t^2-tan^{-1}~t+C$
Note that $s(0) = 1$
We can find the value of $C$:
$s(0) = (0)^2-tan^{-1}~(0)+C= 1$
$0-0+C = 1$
$C = 1$
Therefore:
$s(t) = t^2-tan^{-1}~t+1$
