$$
\begin{aligned}
& \text { } I=\int \frac{d x}{\sin x+\sin 2 x} \\
& =\int \frac{d x}{\sin x(1+2 \cos x)}=\int \frac{\sin x d x}{\sin ^2 x(1+2 \cos x)} \\
& =\int \frac{\sin x d x}{(1-\cos x)(1+\cos x)(1+2 \cos x)}
\end{aligned}
$$
Let $\cos x=t \Rightarrow-\sin x d x=d t$
So,
$$
I=-\int \frac{d t}{(1-t)(1+t)(1+2 t)}
$$
By partial fraction method
$$
\begin{aligned}
& \frac{1}{(1-t)(1+t)(1+2 t)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{1+2 t} \\
& \Rightarrow 1=A(1+t)(1+2 t)+B(1-t) \\
& A=\frac{1}{6}, B=\frac{-1}{2} \text { and } C=\frac{4}{3}
\end{aligned}
$$
$A=\frac{1}{6}, B=\frac{-1}{2} \quad$ and $C=\frac{4}{3}$
So,
$$
\begin{aligned}
I=-\frac{1}{6} \int \frac{d t}{1-t} & +\frac{1}{2} \int \frac{d t}{1+t}-\frac{4}{3} \int \frac{d t}{1+2 t} \\
=\frac{1}{6} \log (1-\cos x) & +\frac{1}{2} \log (1+\cos x) \\
& -\frac{2}{3} \log (1+2 \cos x)+c .
\end{aligned}
$$