The value of the integral $$\int_1^2 \left(\frac{t^4+1}{t^6+1}\right) dt$$ is:

We need to evaluate $$\int_1^2 \frac{t^4+1}{t^6+1} dt$$.

Note that $$t^6 + 1 = (t^2+1)(t^4-t^2+1)$$ so that $$\frac{t^4+1}{t^6+1} = \frac{t^4+1}{(t^2+1)(t^4-t^2+1)}.$$ Since $$t^4 + 1 = (t^4 - t^2 + 1) + t^2$$ it follows that $$\frac{t^4+1}{(t^2+1)(t^4-t^2+1)} = \frac{1}{t^2+1} + \frac{t^2}{(t^2+1)(t^4-t^2+1)}.$$

A cleaner approach is to write $$\frac{t^4+1}{t^6+1} = \frac{1}{1+t^2} + \frac{t^2}{(1+t^2)(t^4-t^2+1)},$$ and since $$(1+t^2)(t^4-t^2+1)=t^6+1$$ the second term is $$\frac{t^2}{t^6+1}$$. Thus $$\int \frac{t^4+1}{t^6+1} dt = \int \frac{1}{1+t^2} dt + \int \frac{t^2}{1+t^6} dt.$$

The first integral is $$\tan^{-1}t$$. For the second, use the substitution $$u = t^3$$, $$du = 3t^2 dt$$, to obtain $$\int \frac{t^2}{1+t^6} dt = \frac{1}{3}\int \frac{du}{1+u^2} = \frac{1}{3}\tan^{-1}(u) + C = \frac{1}{3}\tan^{-1}(t^3) + C.$$ Hence, $$\int \frac{t^4+1}{t^6+1} dt = \tan^{-1}t + \frac{1}{3}\tan^{-1}(t^3) + C.$$

Evaluating from 1 to 2 gives $$[\tan^{-1}t + \frac{1}{3}\tan^{-1}(t^3)]_1^2 = \bigl(\tan^{-1}2 + \frac{1}{3}\tan^{-1}8\bigr) - \bigl(\tan^{-1}1 + \frac{1}{3}\tan^{-1}1\bigr) = \tan^{-1}2 + \frac{1}{3}\tan^{-1}8 - \frac{\pi}{4} - \frac{\pi}{12} = \tan^{-1}2 + \frac{1}{3}\tan^{-1}8 - \frac{\pi}{3}.$$

The final result is $$\boxed{\tan^{-1}2 + \frac{1}{3}\tan^{-1}8 - \frac{\pi}{3}}$$ which corresponds to Option C.