Step 1: By Parts:
Compute $\int\dfrac{dx}{\log x}$ by parts with $u=\dfrac1{\log x}$ and $dv=dx$: it equals $\dfrac x{\log x}+\int\dfrac{dx}{(\log x)^2}$.
Step 2: Subtract:
The derivative of $\dfrac1{\log x}$ is $-\dfrac1{x(\log x)^2}$, and multiplying by $x$ gives $-\dfrac1{(\log x)^2}$. Subtracting $-$(that integral) in the by-parts formula gives the plus sign above. Moving the integral to the left side:
\[ \int\frac{dx}{\log x}-\int\frac{dx}{(\log x)^2}=\frac{x}{\log x}+c \]
Step 3: Answer:
The result is $\dfrac{x}{\log x}+c$. Option (C).
Final Answer:
Option (C).
\[ \boxed{\text{(C) } \frac{x}{\log x}+c} \]