$$\int \frac{\sin{2x}}{\sin{x}+\cos^2{x}}dx=\int \frac{2\sin{x}\cos{x}}{\sin{x}+1-\sin^2{x}}dx=\left| \begin{array}{c} t=\sin x \\ dt=\cos x\,dx \end{array} \right|=\int \frac{2t}{-t^2+t+1}dt$$
Now I see that $2t$ in numerator and $-t^2$ in denominator, i want to do substitution, but $t$ is the way. I am stuck here.
On
$$\int \frac{\sin{2x}}{\sin{x}+\cos^2{x}}dx=\int \frac{2\sin{x}\cos{x}}{\sin{x}+1-\sin^2{x}}dx$$ $$\left| t=\sin x,\quad dt=\cos x\,dx \right|$$ $$=\int \frac{2t}{-t^2+t+1}dt$$ $$=\int\frac{-2t}{t^2 - t -1}dt$$ $$ =-\left(\int \frac{2t - 1}{t^2 - t-1}\,dt + \int \frac {1}{t^2 -t - 1}\,dt\right)\tag{1}$$
For the first integral, put $$u = t^2 - t -1\implies du = 2t - 1$$...and proceed from there.
For the second integral $${t^2 -t-1} = \frac{-1}{4}\left(-2t + \sqrt 5+1 \right)\left(2t + \sqrt 5-1\right) $$
so you can compute the second integral via partial fractions.
On
After the substitution your integral is $$ \int\frac{-2t}{(t-\varphi)(t-\hat\varphi)}\,dt $$ where $$ \varphi=\frac{1+\sqrt{5}}{2}, \qquad \hat\varphi=\frac{1-\sqrt{5}}{2}=1-\varphi $$ so the decomposition into partial factors is $$ \frac{-2t}{(t-\varphi)(t-\hat\varphi)}= \frac{A}{t-\varphi}+\frac{B}{t-\hat\varphi} $$ which results in $$ (A+B)t-(A\hat\varphi+B\varphi)=-2t $$ so $A+B=-2$ and $B=-2-A$. From the constant term we get $$ A\hat\varphi-A\varphi-2\varphi=0 $$ which gives $$ A=\frac{2\varphi}{\hat\varphi-\varphi}=-\frac{2\varphi}{\sqrt{5}} $$ and $$ B=-2-A=\frac{2\hat\varphi}{\sqrt{5}} $$ Thus the integral is $$ \frac{2\hat\varphi}{\sqrt{5}}\log|t-\hat\varphi|- \frac{2\varphi}{\sqrt{5}}\log|t-\varphi|+C $$ Substitute back $t=\sin x$ and you're done.
Rewriting $$\frac{2t}{t^2-t-1}=\frac{2t-1}{t^2-t-1}+\frac{1}{t^2-t-1}$$
We see that the first summand equals the following logarithmic derivative $$(\log(t^2-t-1))'$$ Whereas the second term can be easily dealt with.
Remark: from your question, it seemed that the problem was dealing with the $t$ in the numerator, so I only did that.