Evaluate :

$\displaystyle\int {\dfrac{9cosx-sinx}{4sinx+5cosx}dx}$

$\to \displaystyle \int \dfrac{9\cos x - \sin x}{4\sin x + 5 \cos x} dx$

It is in the form $\dfrac{f(x)}{g(x)}$, so we will write $f(x)$ as $A(g(x)) + B(g'(x))$.

$\to f(x) = 9\cos x - \sin x$

$\to g(x) = 4\sin x + 5 \cos x$

$g'(x) = 4 \cos x - 5 \sin x$

So, $f(x) = g(x) + g'(x)$

$\to \displaystyle \int 1+ \dfrac{g'(x)}{g(x)} dx \Rightarrow x + ln (g(x))+C$

=$x+\ln(4\sin x+5\cos x)+c$