You are dealing with the *geometric* distribution. Let the probability of "success" on any one trial be $p\ne 0$. Suppose you repeat the experiment independently until the first success. The mean waiting time until the first success is $\frac{1}{p}$.

For a proof, let $X$ be the waiting time until the first success, and let $e=E(X)$ There are two possibilities. Either you get success right away (probability $p$) in which case your waiting time is $1$, or you get a failure on the first trial (probability $1-p$), in which case your expected waiting time is $1+e$, since the first trial has been "wasted." Thus
$$e=(1)(p)+(1+e)(1-p).$$
Solving for $e$ we get $e=\frac{1}{p}$.

Alternately, we can set up an infinite series expression for $e$, and sum the series. The probability that $X=1$ is $p$. The probability that $X=2$ is $(1-p)p$ (failure then success). The probability that $X=3$ is $(1-p)^2p$ (two failures, then success). And so on. So
$$E(X)=p+2(1-p)p+3(1-p)^2p +4(1-p)^3p +\cdots.$$

If you wish, I can describe the process of finding a simple expression for the above infinite series.

**Remark:** For smallish $p$, your integral turns out to be a reasonably good approximation to the mean. However, as you are aware, we have here a *discrete* distribution, and the right process involves summation, not integration.

Another tool that we can use is the fact that
$$E(X)=\Pr(X \ge 1)+\Pr(X\ge2)+\Pr(X \ge 3)+\cdots.$$
That is the discrete version of your integral.
It gives us the geometric progression $1+(1-p)+(1-p)^2+(1-p)^3+\cdots$, which has sum $\frac{1}{p}$ if $p\ne 0$.