Step 1: Understand the task.
We are told $\int\dfrac{\cos8x+1}{\cot2x-\tan2x}\,dx=A\cos8x+c$ and must find the number $A$. We will simplify the integrand into something easy.
Step 2: Simplify the denominator.
Write cot and tan as ratios: \[ \cot2x-\tan2x=\frac{\cos2x}{\sin2x}-\frac{\sin2x}{\cos2x}=\frac{\cos^22x-\sin^22x}{\sin2x\cos2x}. \] The top is $\cos4x$ and the bottom is $\tfrac{1}{2}\sin4x$, so the denominator equals $2\cot4x$.
Step 3: Simplify the numerator.
Use $1+\cos8x=2\cos^24x$. So the numerator is $2\cos^24x$.
Step 4: Combine.
\[ I=\int\frac{2\cos^24x}{2\cot4x}\,dx=\int\cos^24x\cdot\frac{\sin4x}{\cos4x}\,dx=\int\cos4x\sin4x\,dx. \]
Step 5: Use a double angle.
Since $\cos4x\sin4x=\tfrac{1}{2}\sin8x$, \[ I=\frac{1}{2}\int\sin8x\,dx=\frac{1}{2}\cdot\left(-\frac{\cos8x}{8}\right)+c=-\frac{1}{16}\cos8x+c. \]
Step 6: Read off $A$.
Comparing with $A\cos8x+c$, we get $A=-\tfrac{1}{16}$.
\[ \boxed{-\dfrac{1}{16}} \]