1. Rewrite the integrand

We need to evaluate

$$\int \frac{5\tan x}{\tan x-2}\,dx.$$

Use $\tan x = \dfrac{\sin x}{\cos x}$:

$$\frac{5\tan x}{\tan x-2} = \frac{5\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}-2} = \frac{5\sin x}{\sin x-2\cos x}.$$

So the integral becomes

$$I=\int \frac{5\sin x}{\sin x-2\cos x}\,dx.$$

2. Match with the given form

We are told

$$I = x + a\ln|\sin x-2\cos x| + k.$$

Differentiate the right-hand side:

$$\frac{d}{dx}\left(x + a\ln|\sin x-2\cos x|\right) = 1 + a\,\frac{\cos x+2\sin x}{\sin x-2\cos x}.$$

This must equal the integrand:

$$1 + a\frac{\cos x+2\sin x}{\sin x-2\cos x} = \frac{5\sin x}{\sin x-2\cos x}.$$

3. Combine into a single fraction

Write $1$ over the same denominator:

$$1 = \frac{\sin x-2\cos x}{\sin x-2\cos x}.$$

Hence

$$\frac{\sin x-2\cos x + a(\cos x+2\sin x)}{\sin x-2\cos x} = \frac{5\sin x}{\sin x-2\cos x}.$$

So the numerators must be equal:

$$\sin x-2\cos x + a\cos x + 2a\sin x = 5\sin x.$$

Group terms:

$$(1+2a)\sin x + (-2+a)\cos x = 5\sin x + 0\cos x.$$

4. Compare coefficients

For $\sin x$:

$$1+2a=5 \implies 2a=4 \implies a=2.$$

For $\cos x$:

$$-2+a=0 \implies a=2.$$

Both agree.

5. Final answer

$$\boxed{a=2}$$

So the correct option is D.