To solve these problems, we will use the Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus states that if $F(x) = \int_{a}^{g(x)} f(t) \, dt$, then the derivative $F'(x) = f(g(x)) \cdot g'(x)$.

Let's go through each part:

Part (a)

Given: $F(x) = \int_{8}^{x} \frac{1}{t} \, dt$ Using the Fundamental Theorem of Calculus: $F'(x) = \frac{1}{x} \cdot \frac{d}{dx}(x) = \frac{1}{x}$

Part (b)

Given: $F(x) = \int_{x}^{17} \frac{1}{t} \, dt$ Here, the upper and lower bounds are switched. We can rewrite this as: $F(x) = -\int_{17}^{x} \frac{1}{t} \, dt$ Then, $F'(x) = -\frac{1}{x} \cdot \frac{d}{dx}(x) = -\frac{1}{x}$

Part (c)

Given: $F(x) = \int_{20}^{x^3} \frac{1}{t} \, dt$ Let $g(x) = x^3$, then $g'(x) = 3x^2$. Applying the Fundamental Theorem of Calculus: $F'(x) = \frac{1}{x^3} \cdot 3x^2 = \frac{3}{x}$

Part (d)

Given: $F(x) = \int_{2 + \cos x}^{x^2 + 1} \frac{1}{t} \, dt$ Let $g(x) = x^2 + 1$ and $h(x) = 2 + \cos x$.

Applying the Fundamental Theorem of Calculus and using the chain rule: $F'(x) = \frac{1}{x^2 + 1} \cdot 2x - \frac{1}{2 + \cos x} \cdot (-\sin x)$ Simplifying: $F'(x) = \frac{2x}{x^2 + 1} + \frac{\sin x}{2 + \cos x}$

Summary of Answers

1. (a) $F'(x) = \frac{1}{x}$
2. (b) $F'(x) = -\frac{1}{x}$
3. (c) $F'(x) = \frac{3}{x}$
4. (d) $F'(x) = \frac{2x}{x^2 + 1} + \frac{\sin x}{2 + \cos x}$

Would you like further details on any of these calculations or have any specific questions?

Here are some related questions to explore:

1. How does the Fundamental Theorem of Calculus apply when the bounds are functions of $x$?
2. What happens if the integrand itself is a function of $x$ rather than just $t$?
3. How can the Fundamental Theorem of Calculus be used in applications of physics, like calculating work done by a force?
4. What are the implications of switching the limits of integration on the integral’s value?
5. How do we handle differentiation under the integral sign in more complex functions?

Tip: Always check if the bounds of an integral are functions of $x$ before differentiating, as this affects the derivative of the integral.