I am going to take some liberties interpreting this question (and its helpful title) as asking,
For constants $C$ and $d$ and uniformly distributed random variable $\epsilon$ supported on $[-1/2,1/2],$ what is the distribution of $Z = C/(d+\epsilon)$?
Before we go on, the Wikipedia article derives the distribution of $1/U$ where $U$ has a uniform distribution on the interval $[0, t]$ with $t\gt 0.$ This is done by computing
$$\Pr\left(\frac{1}{U}\le y\right) = \Pr\left(U \ge \frac{1}{y}\right) = \frac{1}{t}\max\left(0, t - \frac{1}{y}\right) = \max\left(0, 1 - \frac{1}{yt}\right)\tag{*}$$
for all numbers $y\ge 0$ (since obviously $\Pr(1/U\le y)=0$ for all negative $y$). We'll capitalize on this at the very end.
Returning to the problem at hand, applying the same basic definition works best: the value of the distribution function of $Z$ at any number $z$ is
$$F_Z(z) = \Pr(Z \le z) = \Pr\left(\frac{C}{d+\epsilon}\le z\right).$$
To simplify the algebra it would be nice to clear the denominator, but that's made tricky because when it's negative, the inequality is reversed. So, to that end, consider the following:
When $C=0,$ $Z=0$ constantly, solving the problem. From now on then assume $C\ne 0.$
When $C\lt 0,$ the event $C/(d+\epsilon)\le z$ is equivalent to $1/(d+\epsilon)\ge z/C.$ Otherwise, when $C\gt 0,$ the equivalent event is $1/(d+\epsilon)\le z/C.$
The support of the denominator $d+\epsilon$ is the interval $[d-1/2, d+1/2].$ If this is entirely positive or entirely negative, things are relatively simple; but otherwise we need to break the interval into its negative and positive parts. An elegant way to handle this, with a minimum of fuss, is to express the distribution of $d+\epsilon$ as a mixture of a positively supported and negatively supported distribution. (When a variable $X$ has a distribution function $F_X$ with weight $q$ and $Y$ has a distribution function $F_Y$ with weight $p,$ the distribution function of their weighted mixture is $qF_X + pF_Y.$)
Specifically, let $l = \min(0, d-1/2)$ and $u=\max(0, d+1/2)$, so that $[l,u]$ covers all the numbers of possible relevance in this analysis. For any two numbers $a$ and $b,$ let $U(a,b)$ designate a uniform distribution on the interval $(\min(a,b),\max(a,b))$ (so we don't care what order $a$ and $b$ might be in). By checking the three cases $u\lt 0,$ $l\le 0 \le u,$ and $l\gt 0,$ you can readily verify that a weighted mixture of a $U(l,0)$ variable with weight $q = 1/2-d$ and a $U(0,u)$ variable with weight $p=d+1/2$ has the same distribution as $d+\epsilon.$
(Any experts who have been following along might object that $p$ or $q$ can be negative. That's correct -- but the math still works out, because $p+q=1$ and, because we know we're computing a valid distribution function (that of $Z$), we are guaranteed the expressions we are about to derive will give a legitimate distribution function.)
To deal with the uniform distributions on $(l,0]$ when $l\lt 0,$ negate the random variable. This will convert the event $1/X \le z/C$ to $1/(-X)\ge -z/C$ (notice the change in the direction of the inequality) and $1/X \ge z/C$ becomes $1/(-X)\le -z/C.$
This analysis has reduced the problem to finding some probabilities that $1/X\le \pm z/C$ or $1/X\ge \pm z/C$ where $X$ has a uniform distribution on a positive interval anchored at zero -- either $(0,u)$ or $[0,-l].$ Apply the results $(*)$ in the Wikipedia article directly in either case and combine those results using the definition of a weighted mixture. The only challenge is to keep track of the $\pm$ signs and the directions of the inequalities.
If any of this looks tricky or abstract, I recommend plotting the distributions involved in each of the three cases.
