$\int \frac { \cos x + 2 \sin x } { 7 \sin x - 5 \cos x } d x = a x + b \ln | 7 \sin x - 5 \cos x | + c$ then $a+b$ is

$\displaystyle\int \frac{\cos x+2 \sin x}{7 \sin x- 5 \cos x}$

Let $\displaystyle 2 \sin (x)+ \cos (x)= \frac{9}{74}(7 \sin x- 5 \cos x)+\frac{17}{74}(5 \sin x+ 7 \cos x)$

$\displaystyle \Rightarrow \int \frac{9/74(7 \sin x- 5 \cos x)+17/74(5 \sin x+7 \cos x)}{(7 \sin x- 5 \cos x)}dx$

$\displaystyle \Rightarrow \int \dfrac{9}{74}dx+\frac{17}{74}\int \frac{5 \sin x+7 \cos x}{7 \sin x-5 \cos x}dx$

Let $u= 7 sin x- 5 cos x$

$\therefore dx=\dfrac{1}{5\sin x+7\cos x}du$

$\displaystyle\therefore \frac{9}{74}x+\dfrac{17}{74}\int \dfrac{1}{u}du$

$\dfrac{9}{74}x+\dfrac{17}{74}\log(u)$

$\therefore \dfrac{9}{74}x+\dfrac{17}{74}\log(7 \sin x-5 \cos x)+c$

$\therefore a=\dfrac{9}{74}, b=\dfrac{17}{74}$

$\therefore a+b\Rightarrow \dfrac{26}{74}\Rightarrow \dfrac{13}{37}$