91Ó°ÊÓ

Partial fraction decomposition is a technique used to break down complex fractions into simpler, more manageable terms. In this exercise, it's applied to the integrand \( \frac{1}{R(R-1)(R+1)} \). First, we express this as a sum of simpler fractions:

• \( \frac{A}{R} \)
• \( \frac{B}{R-1} \)
• \( \frac{C}{R+1} \)

This step involves finding constants \( A \), \( B \), and \( C \) such that the original fraction equals their sum. By multiplying through by the denominator, we eliminate the fractions and expand to get:
\[ 1 = A(R-1)(R+1) + B(R)(R+1) + C(R)(R-1) \]
We then equate coefficients to solve for \( A \), \( B \), and \( C \). Here, \( A = -1 \), \( B = \frac{1}{2} \), and \( C = \frac{1}{2} \). Partial fraction decomposition is crucial for turning a complex fraction into simpler components that are easier to integrate.

Definite integrals represent the signed area under a curve from one point to another on a given interval. In this problem, we are evaluating the definite integral:
\[ \int_{2}^{3} \frac{dR}{R(R-1)(R+1)} \]
This process involves integrating the function over the interval from \( R = 2 \) to \( R = 3 \). After decomposing the integrand into partial fractions, we integrate each term separately:

• \( \int_{2}^{3} -\frac{1}{R} \, dR \)
• \( \int_{2}^{3} \frac{1/2}{R-1} \, dR \)
• \( \int_{2}^{3} \frac{1/2}{R+1} \, dR \)

The power of definite integrals lies in their ability to give precise values for complex areas, leveraging the antiderivative calculated and evaluating it at the two boundaries. This method simplifies what might otherwise be a very laborious computation.

Logarithmic integration often comes into play when integrating functions of the form \( \frac{1}{x} \) or similar expressions that include simple algebraic fractions. In this exercise, each piece of the decomposed fraction becomes suitable for logarithmic integration:

• \( \int \frac{1}{R} \, dR = \ln |R| + C \)
• \( \int \frac{1}{R-1} \, dR = \ln |R-1| + C \)
• \( \int \frac{1}{R+1} \, dR = \ln |R+1| + C \)

After substitution, this allows the integration to proceed smoothly for each term. For example, integrating \( -\frac{1}{R} \) over the interval leads to \(-\ln |R| \bigg|_{2}^{3} \). Each expression is evaluated using their specific limits:

• \(-\ln(3) + \ln(2)\)
• \(\frac{1}{2}\ln(2)\)
• \(\frac{1}{2}(2\ln(2) - \ln(3))\)

Logarithms simplify manipulation and calculation, especially when integrating fractions that result from partial fraction decomposition.