Answer
\begin{align} r(t)&= \left\langle t-t^2+3,2t^2+1 \right\rangle . \end{align}
Work Step by Step
By integration, we have
\begin{align} r(t)&=\int \frac{dr}{dt} dt\\
&=\int \left\langle1-2t,4t \right\rangle dt\\
&= \left\langle t-t^2+c_1,2t^2+c_2 \right\rangle . \end{align} By the condition $r(0)=\lt3,1\gt$, we get $$3=c_1, \quad 1=c_2$$
Hence, we have \begin{align} r(t)&= \left\langle t-t^2+3,2t^2+1 \right\rangle . \end{align}
