It is a question given in a past prelim paper, so is written correct. I had a brain wave, tell me if you think this would work?
y'(x) = t^3 + 3t , y''(x) = 3t^3 + 3t
Now differentiate y'(x) to get,
y''(x) = (3t^2 + 3)(dt/dx) = 3(t^2 +1)(dt/dx)
Therefore, dt/dx = 1/3
So integrating both sides you get,
t = (1/3)x + C
So for x(1) = 4, 1 = 4/3 + C
Therefore, C = - 1/3
So, x(t) = 3t + 1