Answer
$r(t)=(t^2+3t)i+(1-t) j+(\dfrac{t^3}{3}+1)k$
Work Step by Step
As we are given that $a(t)=2i+2tk$ and $v(0)=3i-j$
Since, $v(t)=\int a(t)$ and $r(t)=\int v(t)$
Thus, $v(t)=\int (2i+2tk) dt$
$\implies v(t)=(2t+3)i-j+t^2k$
Now, $r(t)=\int [(2t+3)i-j+t^2k]=(t^2+3t)i-t j+\frac{t^3}{3}k+c$
Here, $c$ represents a constant of integration.
That is, $r(0)=j+k \implies c= j+k$
$r(t)=\int [(2t+3)i-j+t^2k]=(t^2+3t)i-t j+\frac{t^3}{3}k+j+k$
Hence, $r(t)=(t^2+3t)i+(1-t) j+(\dfrac{t^3}{3}+1)k$