To solve this using the First Part of the Fundamental Theorem of Calculus, we need to take the derivative of the function:

$y = \int_{1}^{3x+5} \frac{t}{1 + t^3} \, dt$

We have an integral with a variable upper limit of integration. According to the First Fundamental Theorem of Calculus, if we have a function of the form:

$y = \int_{a}^{g(x)} f(t) \, dt$

where $a$ is a constant and $g(x)$ is a function of $x$, the derivative of $y$ with respect to $x$ is:

$\frac{dy}{dx} = f(g(x)) \cdot g'(x)$

In our case:

• $f(t) = \frac{t}{1 + t^3}$
• The upper limit is $g(x) = 3x + 5$
• The lower limit is constant (1)

Now, we can apply the rule:

$\frac{dy}{dx} = \frac{(3x + 5)}{1 + (3x + 5)^3} \cdot \frac{d}{dx} (3x + 5)$

The derivative of $3x + 5$ is simply 3. Thus, the derivative of $y$ is:

$\frac{dy}{dx} = \frac{3x + 5}{1 + (3x + 5)^3} \cdot 3$

So, the derivative of the given function is:

$\frac{dy}{dx} = \frac{3(3x + 5)}{1 + (3x + 5)^3}$

Would you like to explore any further details or have questions about the process? Here are some related questions:

1. How does the Fundamental Theorem of Calculus apply to integrals with variable limits?
2. Can you explain the difference between the first and second parts of the Fundamental Theorem of Calculus?
3. How do you handle definite integrals with more complex functions inside the integral?
4. What happens if the limits of integration are both functions of $x$?
5. How do you evaluate integrals with cube powers like $t^3$?

Tip: When dealing with integrals with variable limits, always check the derivative with respect to those limits.