$$\int {{{5\tan x} \over {\tan x - 2}}} dx$$

$$ = \int {{{5{{\sin x} \over {\cos x}}} \over {{{\sin x} \over {\cos x}} - 2}}} \,dx$$

$$ = \int {\left( {{{5\sin x} \over {\cos x}} \times {{\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$

$$ = \int {{{5\,\sin \,x\,dx} \over {\sin x - 2\,\cos x}}} $$

$$ = \int {\left( {{{4\sin x + \sin x + 2\cos x - 2\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$

$$ = \int {{{\left( {\sin x - 2\cos x} \right) + \left( {4\sin x + 2\cos x} \right)} \over {\sin x - 2\cos x}}} \,dx$$

$$ = \int {{{\left( {\sin x - 2\cos x} \right) + 2\left( {\cos x + 2\sin x} \right)} \over {\left( {\sin x - 2\cos x} \right)}}} \,dx$$

$$ = \int {{{\sin x - 2\cos x} \over {\sin x - 2\cos x}}dx + 2\int {\left( {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}} \right)} } \,dx$$

$$ = \int {dx + 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos \,x}}} } \,dx$$

$$ = {I_1} + {I_2}$$ where $${I_1} = \int {dx} $$ and

$${I_2} = 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}\,dx} $$

Put $$\sin x - 2\cos x = t$$

$$ \Rightarrow \left( {\cos x + 2\sin x} \right)dx = dt$$

$$\therefore$$ $${I_2} = 2\int {{{dt} \over t}} = 2\ln \,t + C$$

$$ = 2\,\ln \left( {\sin \,x - 2\cos \,x} \right) + C$$

Hence, $${I_1} + {I_2}$$

$$ = \int {dx + 2\ln \left( {\sin x - 2\cos x} \right) + c} $$

$$ = x + 2\ln \left| {\left( {\sin x - 2\cos x} \right)} \right| + k$$

$$ \Rightarrow a = 2$$