We need to evaluate $$\displaystyle I = \int \frac{\ln x}{x^{2}}\,dx$$ for $$x \gt 0$$.
Step 1: Substitution
Put $$t = \ln x \;\; \Longrightarrow \;\; x = e^{t}, \qquad dx = e^{t}\,dt = x\,dt.$$
Rewrite the integrand using this substitution:
$$I = \int \frac{\ln x}{x^{2}}\,dx
= \int \frac{t}{x^{2}}\,(x\,dt)
= \int \frac{t}{x}\,dt
= \int t\,e^{-t}\,dt,$$
because $$\frac{1}{x} = e^{-t}.$$
Step 2: Integration by parts
For $$\int t\,e^{-t}\,dt,$$ choose
$$u = t \;\; \Longrightarrow \;\; du = dt,$$
$$dv = e^{-t}\,dt \;\; \Longrightarrow \;\; v = -e^{-t}.$$
Using the integration-by-parts formula $$\int u\,dv = uv - \int v\,du,$$ we get
$$\int t\,e^{-t}\,dt
= (-t\,e^{-t}) - \int (-e^{-t})\,dt
= -t\,e^{-t} - \bigl(-e^{-t}\bigr) + C
= -t\,e^{-t} - e^{-t} + C.$$
Step 3: Back-substitution
Replace $$t = \ln x$$ and $$e^{-t} = \dfrac{1}{x}:$$
$$I = -(\ln x)\,\frac{1}{x} - \frac{1}{x} + C
= -\frac{\ln x}{x} - \frac{1}{x} + C.$$
Hence, the required indefinite integral is
$$\boxed{-\dfrac{\ln x}{x} - \dfrac{1}{x} + C}.$$
Therefore, the correct choice is:
Option A which is: $$-\dfrac{\ln x}{x} - \dfrac{1}{x} + C.$$
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