Now I want to convert this data to velocity. How would I do that for each of the specific accelerations at their specific times?
You cannot "convert" a single acceleration data point to a velocity. You have to integrate the acceleration to find the change in velocity. (You can not find the absolute velocity unless you know the velocity at some point in time; you will otherwise only be able to find the change in velocity.)
In one-dimension
$$
a = \frac{dv}{dt}\;,
$$
where $a$ is acceleration, $v$ is velocity, and $t$ is time. (N.b., although you used the symbol "$x$" for time, I am using the more-conventional "$t$" for time.)
Therefore, in one dimension:
$$
\Delta v = v_2 - v_1 = \int_{t_1}^{t_2}a(t) dt
$$
The integral can be discretized like:
$$
\int_{t_1}^{t_2} a(t)dt \to \sum_{n} a_n {\delta t}_n\;,
$$
where ${\delta t}_n$ is your time step and $a_n$ is the acceleration now indexed at discrete time points.
With the discretized version, we have:
$$
\Delta v = v_2 - v_1 \approx \sum_{n=n_1}^{n_2-1} a_n {\delta t}_n
=\delta t\sum_{n=n_1}^{n_2-1} a_n\;,
$$
where the last equal sign holds for evenly-spaced data points, and where the sum over $n$ runs over the points corresponding to the first time through the last time, inclusive of the first time, but not the last time (including the first but not the last is is a choice, a different choice could also be made). This is illustrated with examples below.
Update:
For example, denote the velocity at $t=0$ by $v_0$. Then the velocity at $t=0.25$ seconds can be estimated as:
$$
v(t=0.25) = v_0 + a_0 \delta t
$$
$$
= v_0 + (-9.81 \times 0.25)
$$
For example, the velocity at $t=1$ second can be estimated as:
$$
v(t=1) = v_0 + a_0 \delta t + a_{0.25}\delta t + a_{0.5}\delta t + a_{0.75}\delta t
$$
$$
= v_0 + ((-9.81 - 3.93 + 7.84 + 4.31)\times 0.25)
$$
