91Ó°ÊÓ

Partial fraction decomposition is a technique used to break down complex rational functions into simpler ones. This makes them easier to integrate, especially when dealing with polynomial denominators. For the given function, \( \frac{1}{x(x+1)} \), the goal is to express it as a sum of simpler fractions.

The decomposition starts by assuming a format such as:

• \( \frac{A}{x} + \frac{B}{x+1} \)

Here, \( A \) and \( B \) are constants that need to be determined. This transformation allows easy integration of the function. You will solve for \( A \) and \( B \) by expanding and collecting like terms, then equating it to the original numerator.

• In the problem, \( A = 1 \) and \( B = -1 \).
• Verifiable by substituting the values back.

This method simplifies the integration of otherwise complicated functions.

An antiderivative of a function is a function whose derivative is the original function. In the context of definite integrals, finding the antiderivative is a vital step. Here, once we break the original function into simpler parts using partial fraction decomposition, each part is integrated separately.

The integrals of \( \frac{1}{x} \) and \( \frac{1}{x+1} \) are straightforward:

• The antiderivative of \( \frac{1}{x} \) is \( \ln|x| \).
• The antiderivative of \( \frac{1}{x+1} \) is \( \ln|x+1| \).

These results are derived from the natural logarithm function because the derivative of \( \ln|x| \) is \( \frac{1}{x} \). By summing up the antiderivatives of all parts, you'll have the necessary function to apply the limits of integration.

The limits of integration define the interval over which the definite integral is evaluated. In this problem, the function is integrated from \(x = 1\) to \(x = 2\).

To calculate this, you will substitute these limits into the antiderivative you've found. Here is how it works:

• Calculate the antiderivative value at the upper limit (\( x=2 \)).
• Calculate the antiderivative value at the lower limit (\( x=1 \)).
• Subtract the lower limit value from the upper limit value.

This process provides the area under the curve between the specified limits, revealing the definite value for the integral. In our case, the expression simplifies to \( 2\ln2 - \ln3 \). This systematic approach is robust for solving areas under curves in calculus.