Step 1: Simplify the denominator.
We must integrate $\int\dfrac{dx}{\sin x+\sin2x}$. Since $\sin2x=2\sin x\cos x$, the denominator becomes \[ \sin x+2\sin x\cos x=\sin x(1+2\cos x). \]
Step 2: Multiply by $\sin x$ on top and bottom.
\[ I=\int\frac{\sin x\,dx}{\sin^2x(1+2\cos x)}. \] Now replace $\sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x)$.
Step 3: Substitute $u=\cos x$.
Then $du=-\sin x\,dx$, so $\sin x\,dx=-du$. The integral turns into \[ I=-\int\frac{du}{(1-u)(1+u)(1+2u)}. \] This is now a pure algebra problem.
Step 4: Break into partial fractions.
Write \[ \frac{1}{(1-u)(1+u)(1+2u)}=\frac{A}{1-u}+\frac{B}{1+u}+\frac{C}{1+2u}. \] Plugging in $u=1$ gives $A=\tfrac{1}{6}$; $u=-1$ gives $B=\tfrac{1}{2}$; $u=-\tfrac{1}{2}$ gives $C=-\tfrac{4}{3}$.
Step 5: Integrate each piece.
With the outer minus sign folded in, the $\tfrac{1}{1-u}$ term integrates to $+\tfrac{1}{6}\log|1-u|$, the $\tfrac{1}{1+u}$ term to $\tfrac{1}{2}\log|1+u|$, and the $\tfrac{1}{1+2u}$ term to $-\tfrac{2}{3}\log|1+2u|$.
Step 6: Put $u=\cos x$ back.
So \[ I=\frac{1}{6}\log|1-\cos x|+\frac{1}{2}\log|1+\cos x|-\frac{2}{3}\log|1+2\cos x|+c. \]
\[ \boxed{\dfrac{1}{6}\log|1-\cos x|+\dfrac{1}{2}\log|1+\cos x|-\dfrac{2}{3}\log|1+2\cos x|+c} \]