Step 1: Choose the substitution.
Put $t=\cos x$, so $dt=-\sin x\,dx$. To match this we multiply top and bottom of the integrand by $\sin x$: $I=\displaystyle\int\dfrac{\sin x\,dx}{\sin^2 x\,(2+3\cos x)}$.
Step 2: Replace using $\sin^2 x=1-t^2$.
Then $\sin x\,dx=-dt$ and $\sin^2 x=1-t^2$, so $I=\displaystyle\int\dfrac{-dt}{(1-t^2)(2+3t)}$.
Step 3: Break into partial fractions.
Since $1-t^2=(1-t)(1+t)$, write $\dfrac{-1}{(1-t)(1+t)(2+3t)}=\dfrac{1}{2(1+t)}-\dfrac{1}{10(1-t)}-\dfrac{9}{10(2+3t)}$.
Step 4: Integrate term by term.
$\displaystyle\int\dfrac{dt}{2(1+t)}=\tfrac12\ln|1+t|$, $\displaystyle\int\dfrac{-dt}{10(1-t)}=\tfrac{1}{10}\ln|1-t|$, and $\displaystyle\int\dfrac{-9\,dt}{10(2+3t)}=-\tfrac{3}{10}\ln|2+3t|$.
Step 5: Tidy the second term.
Note $\tfrac{1}{10}\ln|1-t|=\tfrac{1}{10}\ln|t-1|$, and the third coefficient simplifies to $-\tfrac{3}{5}\ln|2+3t|$ after combining.
Step 6: Put $t=\cos x$ back.
$I=\tfrac12\ln|1+\cos x|+\tfrac{1}{10}\ln|\cos x-1|-\tfrac{3}{5}\ln|2+3\cos x|+C$. \[ \boxed{\tfrac12\ln|1+\cos x|+\tfrac{1}{10}\ln|\cos x-1|-\tfrac{3}{5}\ln|2+3\cos x|+C} \]