91Ó°ÊÓ

The function given is: $$\int\left(3 t^{2}+\sec ^{2} 2 t\right) d t$$ We can split this integral into two separate integrals: $$\int 3t^2 dt + \int \sec^2 2t dt$$

Now, let's find the indefinite integral of each part: For the first integral, use the power rule for integration: $$\int 3t^2 dt = 3\int t^2 dt = 3\frac{1}{3}t^3 = t^3 + C_1$$ where \(C_1\) is the constant of integration for the first integral. For the second integral, we need to use the substitution method: Let \(u = 2t\). Then, \(du = 2dt\), and \(dt = \frac{1}{2} du\). So our integral becomes: $$\int \sec^2(u) \frac{1}{2} du$$ And the integral of \(\sec^2(u)\) is \(\tan(u)\), so the indefinite integral is: $$\frac{1}{2}\tan(u) + C_2 = \frac{1}{2}\tan(2t) + C_2$$ where \(C_2\) is the constant of integration for the second integral. Now, we can add both integrals together: $$\int\left(3 t^{2}+\sec ^{2} 2 t\right) d t = t^3 + \frac{1}{2}\tan(2t) + C$$ where \(C = C_1 + C_2\).

Now we will differentiate our indefinite integral to check our work. Using the sum rule and chain rule, we have: $$\frac{d}{dt}\left(t^3 + \frac{1}{2}\tan(2t) + C\right) = \frac{d}{dt}(t^3) + \frac{d}{dt}\left(\frac{1}{2}\tan(2t)\right) + \frac{d}{dt}(C)$$ Differentiating each term: $$\frac{d}{dt}(t^3) = 3t^2$$ $$\frac{d}{dt}\left(\frac{1}{2}\tan(2t)\right) = \frac{1}{2}(2\sec^2(2t)) = \sec^2(2t)$$ $$\frac{d}{dt}(C) = 0$$ Add them all together: $$3t^2 + \sec^2(2t) + 0 = 3t^2 + \sec^2(2t)$$ The result matches the initial function, which confirms that our indefinite integral is correct.