= Solution
For $s>0$, integrate the <probability density function> to obtain the <cumulative distribution function>
$$
F_X(x)=\frac12\left[1+\tanh\!\left(\frac{x-\mu}{2s}\right)\right].
$$
If $\mu>0$, the requested <probability> is $F_X(\mu)-F_X(0)$. Since the <hyperbolic tangent> is odd, this gives
$$
\boxed{\mathbb P(0<X<\mu)=
\begin{cases}
\dfrac12\tanh\!\left(\dfrac{\mu}{2s}\right),&\mu>0,\\
0,&\mu\leq0.
\end{cases}}
$$
The second case matters because the question places no sign restriction on $\mu$: the event is empty when its upper endpoint is nonpositive. Using an oriented <integral> from $0$ to a negative $\mu$ would give a negative number, not a <probability>.
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