To evaluate $G(x) = x \cdot \int_{x^3 x}^{e^{3x}} \sin(t^2 + 1) \, dt$, let's break down the expression and solve it step-by-step.

Step-by-Step Solution

1. Understanding the Integral: The function $G(x)$ involves an integral: $\int_{x^3 x}^{e^{3x}} \sin(t^2 + 1) \, dt$ which represents the definite integral of $\sin(t^2 + 1)$ from $t = x^3 x$ to $t = e^{3x}$.

2. Applying the Fundamental Theorem of Calculus: While it's challenging to find a closed-form antiderivative for $\sin(t^2 + 1)$, we can focus on differentiating $G(x)$ if that’s of interest. For differentiation purposes, we use the Leibniz rule for differentiation under the integral sign.

3. Differentiating $G(x)$ with Respect to $x$: Let’s differentiate $G(x)$ as follows: $G(x) = x \cdot \int_{x^3 x}^{e^{3x}} \sin(t^2 + 1) \, dt.$ Using the product rule, we find $G'(x)$ by differentiating each part of the product.

4. Differentiating the Integral Term: According to the Leibniz rule, $\frac{d}{dx} \int_{u(x)}^{v(x)} f(t) \, dt = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x).$ Here, $u(x) = x^3 x$ and $v(x) = e^{3x}$, so $\frac{d}{dx} \int_{x^3 x}^{e^{3x}} \sin(t^2 + 1) \, dt = \sin((e^{3x})^2 + 1) \cdot \frac{d}{dx}(e^{3x}) - \sin((x^3 x)^2 + 1) \cdot \frac{d}{dx}(x^3 x).$

5. Combining the Results: By substituting the derivatives and expressions, we can compute $G'(x)$.

Would you like a detailed evaluation of $G'(x)$ or a simplification for specific values of $x$?

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1. What is the value of $G(x)$ for a specific $x$?
2. How can we approximate the integral numerically?
3. What happens if we replace $\sin(t^2 + 1)$ with a different function?
4. What is the impact of different integral bounds?
5. How does $G(x)$ behave as $x \to \infty$?

Tip: When dealing with complex integrals, consider numerical methods for approximate evaluation if no closed-form solution exists.