Question:

Step 1: Decompose the integrand using algebraic manipulation. First, factor the denominator $t^6+1$ as a sum of cubes: $$t^6+1 = (t^2)^3+1^3 = (t^2+1)( (t^2)^2 - t^2 \cdot 1 + 1^2 ) = (t^2+1)(t^4-t^2+1)$$ Now, rewrite the numerator $t^4+1$ as $(t^4-t^2+1) + t^2$. Substitute these into the integrand: $$\dfrac{t^4+1}{t^6+1} = \dfrac{(t^4-t^2+1)+t^2}{(t^2+1)(t^4-t^2+1)}$$ Split this into two fractions: $$\dfrac{t^4-t^2+1}{(t^2+1)(t^4-t^2+1)} + \dfrac{t^2}{(t^2+1)(t^4-t^2+1)}$$ Simplify each term: $$ \dfrac{1}{t^2+1} + \dfrac{t^2}{t^6+1} $$ Thus, the original integral can be written as the sum of two integrals. Step 2: Split the original integral into two separate integrals. Based on the decomposition from Step 1, the given integral can be written as: $$ \int_1^2\left(\dfrac{t^4+1}{t^6+1}\right)dt = \int_1^2 \dfrac{1}{t^2+1} dt + \int_1^2 \dfrac{t^2}{t^6+1} dt $$ Step 3: Evaluate the first integral. The first integral is a standard form: $$ \int_1^2 \dfrac{1}{t^2+1} dt = \left[\tan^{-1}t\right]_1^2 $$ Evaluate at the limits: $$ \tan^{-1}2 - \tan^{-1}1 $$ Since $\tan^{-1}1 = \dfrac{\pi}{4}$: $$ \tan^{-1}2 - \dfrac{\pi}{4} $$ Step 4: Evaluate the second integral using substitution. Consider the second integral: $$ \int_1^2 \dfrac{t^2}{t^6+1} dt $$ Let $u = t^3$. Then, differentiate $u$ with respect to $t$: $$ du = 3t^2 dt \implies t^2 dt = \dfrac{1}{3} du $$ Next, change the limits of integration: When $t=1$, $u=1^3=1$. When $t=2$, $u=2^3=8$. Substitute $u$ and $du$ into the integral: $$ \int_1^8 \dfrac{1}{(u^2)+1} \left(\dfrac{1}{3} du\right) = \dfrac{1}{3} \int_1^8 \dfrac{1}{u^2+1} du $$ This is also a standard form: $$ \dfrac{1}{3} \left[\tan^{-1}u\right]_1^8 $$ Evaluate at the new limits: $$ \dfrac{1}{3} (\tan^{-1}8 - \tan^{-1}1) $$ Substitute $\tan^{-1}1 = \dfrac{\pi}{4}$: $$ \dfrac{1}{3} \left(\tan^{-1}8 - \dfrac{\pi}{4}\right) $$ Step 5: Combine the results of the two integrals. Add the results from Step 3 and Step 4: $$ \left(\tan^{-1}2 - \dfrac{\pi}{4}\right) + \dfrac{1}{3} \left(\tan^{-1}8 - \dfrac{\pi}{4}\right) $$ Distribute $\dfrac{1}{3}$ and combine terms: $$ \tan^{-1}2 - \dfrac{\pi}{4} + \dfrac{1}{3}\tan^{-1}8 - \dfrac{\pi}{12} $$ Group the constant terms: $$ \tan^{-1}2 + \dfrac{1}{3}\tan^{-1}8 - \left(\dfrac{\pi}{4} + \dfrac{\pi}{12}\right) $$ Find a common denominator for the $\pi$ terms: $$ \tan^{-1}2 + \dfrac{1}{3}\tan^{-1}8 - \left(\dfrac{3\pi}{12} + \dfrac{\pi}{12}\right) $$ $$ \tan^{-1}2 + \dfrac{1}{3}\tan^{-1}8 - \dfrac{4\pi}{12} $$ Simplify the constant term: $$ \tan^{-1}2 + \dfrac{1}{3}\tan^{-1}8 - \dfrac{\pi}{3} $$ Step 6: State the final answer and match with the given options. The value of the integral is $\tan^{-1}2 + \dfrac{1}{3}\tan^{-1}8 - \dfrac{\pi}{3}$. This matches Option 3. The final answer is $\boxed{\text{\tan^{-1}2+\dfrac{1}{3}\tan^{-1}8-\dfrac{\pi}{3}}}$.