Step 1: Understanding the Concept:
The integrand involves a sum of terms and a product of terms. This structure specifically resembles the logarithmic derivative formula: \( \frac{d}{dx} \ln(P) = \frac{P'}{P} \), or directly \( P' = P \cdot \sum \frac{u_i'}{u_i} \).
Step 2: Key Formula or Approach:
1. Let \( P(x) = \prod_{r=1}^{n} (x^2 + r^2) \).
2. Find \( \frac{dP}{dx} \).
Step 3: Detailed Explanation:
Let \( P(x) = (x^2+1^2)(x^2+2^2)\dots(x^2+n^2) \).
Using the product rule for differentiation (or logarithmic differentiation):
\[ \frac{d}{dx} P(x) = P(x) \left[ \frac{2x}{x^2+1^2} + \frac{2x}{x^2+2^2} + \dots + \frac{2x}{x^2+n^2} \right] \]
\[ \frac{dP}{dx} = 2 \cdot P(x) \sum_{r=1}^{n} \frac{x}{x^2+r^2} \]
The integral in the question is:
\[ I = \int_{0}^{1} \left( \sum \frac{x}{x^2+r^2} \right) P(x) \, dx = \int_{0}^{1} \frac{1}{2} \frac{dP}{dx} \, dx \]
\[ I = \frac{1}{2} \int_{0}^{1} dP = \frac{1}{2} [P(1) - P(0)] \]
Evaluate \( P(1) \) and \( P(0) \):
\( P(1) = \prod_{r=1}^{2013} (1+r^2) \)
\( P(0) = \prod_{r=1}^{2013} (0^2+r^2) = 1^2 \cdot 2^2 \cdot 3^2 \dots (2013)^2 = ((2013)!)^2 \)
Comparing with the given form \( \frac{1}{2} [\prod (1+r^2) - K] \):
\( K = P(0) = ((2013)!)^2 \).
Step 4: Final Answer:
By identifying the integrand as the derivative of the product, we find \( K = ((2013)!)^2 \).