I'll assume that's an exponent at the end there:
[tex]x(t) = (t+1)(t-3)^3[/tex]
The first derivative gives the velocity. The second derivative gives the acceleration; increasing velocity is the same as positive acceleration. So we want to find when the second derivative is positive.
Let's see if we can use [tex]d(uv) = u\, dv + v\, du[/tex] to avoid multiplying this out.
[tex]x'(t) = 3 (t+1)(t-3)^2 + (t-3)^3[/tex]
That worked; let's do it again:
[tex]x''(t) = 3( 2(t+1)(t-3) + (t-3)^2) + 3(t - 3)^2 [/tex]
[tex]x''(t) = 3(t-3) ( 2(t+1) + (t-3) + (t - 3) ) = 3(t-3)(4t-4)=12(t-1)(t-3)[/tex]
That's a nice parabola. It's zero or negative for [tex]1 \le t \le 3[/tex] so positive everywhere else:
Answer: Increasing velocity when t < 1 or t > 3
