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We are given that \( G(x) = \int_{h_1(x)}^{h_2(x)} f(t) \, dt \), where:
- \( h_1(x) = \pi \cos(2\pi x) \)
- \( h_2(x) = 4 \tan^{-1}(x) \)
- \( f(t) = \frac{188}{1 + t^2} \)

We need to find \( G(1) + G'(1) \).

### Step 1: Find \( G(1) \)

To compute \( G(1) \), we need to evaluate the definite integral for \( x = 1 \).

- \( h_1(1) = \pi \cos(2\pi \cdot 1) = \pi \cos(2\pi) = \pi \cdot 1 = \pi \)
- \( h_2(1) = 4 \tan^{-1}(1) = 4 \cdot \frac{\pi}{4} = \pi \)

Therefore, \( G(1) = \int_{\pi}^{\pi} \frac{188}{1 + t^2} \, dt = 0 \), because the limits of integration are equal.

### Step 2: Differentiate \( G(x) \) to find \( G'(x) \)

By the Leibniz rule for differentiating integrals with variable limits, we have:
\[
G'(x) = f(h_2(x)) \cdot h_2'(x) - f(h_1(x)) \cdot h_1'(x)
\]
where \( f(t) = \frac{188}{1 + t^2} \), \( h_1(x) = \pi \cos(2\pi x) \), and \( h_2(x) = 4 \tan^{-1}(x) \).

#### Step 2.1: Compute \( h_1'(x) \) and \( h_2'(x) \)

- \( h_1(x) = \pi \cos(2\pi x) \)
\[
h_1'(x) = -2\pi^2 \sin(2\pi x)
\]

- \( h_2(x) = 4 \tan^{-1}(x) \)
\[
h_2'(x) = \frac{4}{1 + x^2}
\]

#### Step 2.2: Evaluate \( G'(1) \)

Now, evaluate each term at \( x = 1 \):

- \( h_1(1) = \pi \)
- \( h_2(1) = \pi \)
- \( f(h_1(1)) = f(\pi) = \frac{188}{1 + \pi^2} \)
- \( f(h_2(1)) = f(\pi) = \frac{188}{1 + \pi^2} \)
- \( h_1'(1) = -2\pi^2 \sin(2\pi) = 0 \)
- \( h_2'(1) = \frac{4}{1 + 1^2} = 2 \)

Thus,
\[
G'(1) = f(\pi) \cdot 2 - f(\pi) \cdot 0 = 2 \cdot \frac{188}{1 + \pi^2}
\]

### Step 3: Final Answer

Now we can compute \( G(1) + G'(1) \):

\[
G(1) + G'(1) = 0 + 2 \cdot \frac{188}{1 + \pi^2} = \frac{376}{1 + \pi^2}
\]

This is the required result.

Would you like more details, or have any further questions?

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Here are 5 related questions for practice:
1. How does the Leibniz rule apply to integrals with variable limits?
2. How would \( G'(x) \) change if \( h_1(x) \) and \( h_2(x) \) were linear functions?
3. What is the geometric interpretation of the function \( G(x) \)?
4. How do you find the derivative of an inverse trigonometric function like \( \tan^{-1}(x) \)?
5. How does changing the function \( f(t) \) affect the value of \( G(1) \)?

**Tip:** When differentiating definite integrals with variable limits, always remember to apply the Leibniz rule, accounting for both upper and lower limits of integration!

Let G(x) = ∫ from h1(x) to h2(x) f(t) dt, where h1(x) = π cos(2πx), h2(x) = 4tan⁻¹(x), and f(t) = 188/(1 + t²). Find G(1) + G'(1).

Learn how to calculate G(1) + G'(1) using the Leibniz rule for differentiating integrals with variable limits. This problem involves key concepts such as the Leibniz rule, integral calculus, and inverse trigonometric functions. Detailed step-by-step solution included.

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