Is the moon's revolution speed increasing or decreasing I've come across multiple answers to this question. So, the moon is moving away from the Earth a couple of centimeters per year and the Earth's rotation is slowing down. Here it says the moon's revolution is speeding up, another place it says it is slowing down. So, due to tidal effects, is the moon's revolution slowing down or speeding up?
From the cited article,


  The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth.


I suspect that this is the statement that is causing the OP to be confused, and rightly so. This statement is incorrect. Whether "speed of rotation" means orbital velocity or angular velocity, both are decreasing as the Moon slowly retreats from the Earth.

The Moon is retreating very slowly from the Earth, currently about 3.78 cm per year. Dividing by 385000 km (the Earth-Moon mean orbital radius) yields about 10-10 parts per year. That qualifies as "very slowly". For simplicity, I'll assume the Moon's orbit is essentially circular, plus this very slow spiraling out. For a circular orbit, the relation between orbital radius and angular velocity is $$r^3\omega^2 = G(M+m)$$
This essentially is Kepler's third law. Assuming $G$, $M$, and $m$ are constant, differentiating both sides with respect to time yields
$$r^2\omega (3\dot r \omega + 2 r \dot \omega) = 0$$
meaning
$$r \dot \omega = -\frac32 \dot r \omega$$
Suppose some quantity $q$ is of the form to $q=r^n\omega^m$. Differentiating with respect to time yields $\dot q = r^{n-1}\omega^{m-1}\left(n\dot r\omega + m r \dot\omega\right)$. Since $r\dot\omega = -\frac32 \dot r \omega$, this can be rewritten as $$\dot q = \left(n - \frac 3 2 m\right) \dot r r^{n-1}\omega^m$$
Quantities of this form include angular velocity ($q=\omega$, so $n=0, m=1$ in this case), orbital velocity $(q=r\omega$, so $n=m=1$), and angular momentum (q$=r^2\omega$, so $n=2, m=1$).  Note that $(n-\frac 32m)$ is negative for angular and orbital velocity but positive for orbit angular momentum. Orbit angular momentum thus has the same sign as does $\dot r$ while angular velocity and orbital velocity have the opposite sign as does $\dot r$.

As a sanity check, look to centripetal acceleration. This is given by $a = r\omega^2$. In terms of the above, $n=1$ and $m=2$. The term $n-\frac32m$ is clearly negative in this case, so centripetal acceleration decreases as the Moon retreats from the Earth. Centripetal acceleration is also give by Newton's law of gravitation: It is proportional to the inverse of the square of the distance between the Earth and the Moon, so this does indeed decrease as the Moon retreats from the Earth.