Step 1: Turn everything into exponentials.
Recall $\cosh x=\dfrac{e^x+e^{-x}}{2}$ and $\sinh x=\dfrac{e^x-e^{-x}}{2}$, so $\cosh x+\sinh x=e^x$.
Step 2: Simplify the right-hand side.
The equation $4+6(e^{2x}+1)\tanh x=11\cosh x+11\sinh x$ becomes \[ 4+6(e^{2x}+1)\tanh x=11e^x. \] Step 3: Simplify the tanh term.
Since $\tanh x=\dfrac{e^{2x}-1}{e^{2x}+1}$, we have $(e^{2x}+1)\tanh x=e^{2x}-1$.
Step 4: Substitute back.
The equation becomes \[ 4+6(e^{2x}-1)=11e^x, \] that is $6e^{2x}-2=11e^x$.
Step 5: Solve the quadratic in $e^x$.
Let $t=e^x$. Then $6t^2-11t-2=0$. Factoring, $(6t+1)(t-2)=0$, so $t=2$ or $t=-\dfrac{1}{6}$. Since $e^x>0$, we keep $t=2$.
Step 6: Recover $x$.
From $e^x=2$ we get $x=\log 2$.
\[ \boxed{\log 2} \]