Integral $\int\frac{\sin2x+2tanx}{\cos^{6}x+6\cos^{2}x+4}dx$ Q) $\int\frac{\sin2x+2tanx}{\cos^{6}x+6\cos^{2}x+4}dx$
Tried to split it and integrate individual but , but getting an irreducible cubic equation in denominator . 
Any magic substitution ?

The substitution $u=\tan(x/2)$ will give a rational function, which in principle could be integrated. Here is another solution, that we are lucky to do. One could also in this case do $u=\tan x$ directly...
Using that
$$
\sin 2x=2\sin x\cos x=2\tan x\cos^2 x=\frac{2\tan x}{1+\tan^2x}
$$
and
$$
\cos^6x+6\cos^2x+4=\frac{1+6(1+\tan^2x)+4(1+\tan^2x)^3}{(1+\tan^2x)^3},
$$
we get, after some expanding,
$$
\frac{1}{12}\int \frac{48\tan x+72\tan^3x+24\tan^5x}{11+24\tan^2x+18\tan^4x+4\tan^6x}(1+\tan^2x)\,dx.
$$
Here, we are really lucky(?). Since
$$
D(11+24u^2+18u^4+4u^6)=48u+72u^3+24u^5,
$$
and since $D\tan x=1+\tan^2x$, we find that
$$
\begin{gathered}
\frac{1}{12}\int \frac{48\tan x+72\tan^3x+24\tan^5x}{11+24\tan^2x+18\tan^4x+4\tan^6x}(1+\tan^2x)\,dx\\
=\frac{1}{12}\log(11+24\tan^2x+18\tan^4x+4\tan^6x)+C.
\end{gathered}
$$
