Finding optimal decision threshold for binary comm system
Let P(0) ( P(1) ) is a probability of transmitting bit zero (one); P(e|0) ( P(e|1) ) is a probability of error when detecting bit zero (one).
The probability of erroneous detection is
$$
P(e) = P(e|0)P(0) + P(e|1)P(1) \tag {1}
$$

Let $V_0$ ( $V_1$ ) be a nominal signal voltage of bit zero ( one ) signal at the transmitter.
$$
P(e|0) = \int_T^{\infty}{{\frac {1} {\sigma \sqrt{2\pi}}}\exp\left(-(\nu-V_0)^2/2 {\sigma}^2\right) d\nu } \\
P(e|1) = \int_{-\infty}^T{{\frac {1} {\sigma \sqrt{2\pi}}}\exp\left(-(\nu-V_1)^2/2 {\sigma}^2\right) d\nu }
$$
where $T$ is a detection threshold (OP's $\mu$).

Differentiating $P(e)$ of eq.1 w.r.t. $T$, we arrive at
$$
-P(0){{\frac {1} {\sigma \sqrt{2\pi}}}\exp\left(-(T-V_0)^2/2 {\sigma}^2\right)} + P(1){{\frac {1} {\sigma \sqrt{2\pi}}}\exp\left(-(T-V_1)^2/2 {\sigma}^2\right)} \tag {2}
$$
To find an optimal threshold, we equate the expression in eq.2 to zero:
$$
P(0)\exp\left(-{\frac {(T-V_0)^2} {2 {\sigma}^2}}\right) = P(1)\exp\left(-{\frac {(T-V_1)^2} {2 {\sigma}^2}}\right) \\
T = {\frac {V_0+V_1} {2}} + {\sigma}^2{\ln{(\frac {P(1)} {P(0)})}}
$$
that is, exactly the OP's  result.
The OP's reference source may compute the optimal threshold for the system in which the probabilities of zero/one bit transmissions are equal. The variance ${\sigma} ^2$ is multipled by $\ln{(\frac {P(1)} {P(0)})}$, and it disappears from the expression for optimal threshold, when the zero/one probabilities in transmission are equal ($P(0)=P(1)=1/2$).