$\frac{a+b^{22}}{a^2+b^{11}}$

Let $\log_ab=x.$ Then $\log_ba=\frac{1}{x}$ Rearrange the given information. $x+\dfrac{6}{11x}=\dfrac{67}{11}$ Multiplying by $11x,$ $11x^2+6=67x,$ so $11x^2-67x+6=0.$ Solving this equation for $x$ yields that $x=6$ or $x=\frac{1}{11}.$ Since $x=\log_ab$ and $a>b,$ $\log_ab<1.$ This causes $x=6$ to be an extraneous solution, so $x=\log_ab=\frac{1}{11}.$ This means that $b=\sqrt[11]{a}.$ Now plug this in to the expression we are trying to find. $\dfrac{a+b^{22}}{a^2+b^{11}}=\dfrac{a+(\sqrt[11]{a})^{22}}{a^2+(\sqrt[11]{a})^{11}}=\dfrac{a+a^2}{a^2+a}=\boxed{1}$

$\log_{a}{b}+\frac{6}{11}\log_{b}{a}=\frac{67}{11}$

$\log_{a}{b}+\frac{6}{11}\left(\frac{1}{\log_{a}{b}}\right)=\frac{67}{11}$

$\text{ Let }x=\log_{a}{b}$

$x+\frac{6}{11x}=\frac{67}{11}$

$\text{Multiply both sides by }11x$

$11{x}^{2}+6=67x$

$11{x}^{2}-67x+6=0$

$(11x-1)(x-6)=0$

$x=\frac{1}{11} ;x=6$

$x=\log_{a}b=\frac{1}{11}; a^{\frac{1}{11}}=b; \boxed{a=b^{11}}$

$x=\log_{a}b=6;{a}^{6}=b$

$\text{Since a is greater than b, we disregard the }{a}^{6}=b\text{ because both must be greater than }1.$

$\text{Solve for }\displaystyle\frac{a+{b}^{22}}{{a}^{2}+{b}^{11}}.$

$\displaystyle\frac{{b}^{11}+{b}^{22}}{{{(b}^{11})}^{2}+{b}^{11}}=\frac{{b}^{11}+{b}^{22}}{{b}^{22}+{b}^{11}}=1$