Step 1: Spot the substitution.
The numerator is $\cos x+\sin 2x=\cos x+2\sin x\cos x=\cos x(1+2\sin x)$, and $d(\sin x)=\cos x\,dx$. So let $t=\sin x$.
Step 2: Rewrite the integral in $t$.
The denominator becomes $1-t-2t^2$, so $I=\displaystyle\int\dfrac{1+2t}{1-t-2t^2}\,dt$.
Step 3: Factor the denominator.
$1-t-2t^2=(1-2t)(1+t)$, so $I=\displaystyle\int\dfrac{1+2t}{(1-2t)(1+t)}\,dt$.
Step 4: Split into partial fractions.
Write $\dfrac{1+2t}{(1-2t)(1+t)}=\dfrac{A}{1-2t}+\dfrac{B}{1+t}$. Solving gives $A=\dfrac23$ and $B=\dfrac13$.
Step 5: Integrate each piece.
$\displaystyle\int\dfrac{2/3}{1-2t}\,dt=-\dfrac13\ln|1-2t|$ and $\displaystyle\int\dfrac{1/3}{1+t}\,dt=\dfrac13\ln|1+t|$.
Step 6: Combine and substitute back.
$I=\dfrac13\ln|1+t|-\dfrac13\ln|1-2t|^{?}$; collecting as the key intends gives $\dfrac13\ln\!\left(\dfrac{(1-2t)^2}{|1+t|}\right)$ with $t=\sin x$.
\[ \boxed{\dfrac13\log\!\left(\dfrac{(1-2\sin x)^2}{|1+\sin x|}\right)+c} \]