91Ó°ÊÓ

When dealing with rational functions, often the complexity can be reduced through a technique called partial fraction decomposition. This is particularly useful in calculus, especially when integrating. Let's apply this to the function \( \frac{2}{x^2 - 1} \). Here, the denominator can be factored into \((x-1)(x+1)\).
By expressing \( \frac{2}{x^2 - 1} \) as a sum of two simpler fractions, we write:

• \( \frac{2}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1} \)

This means solving for \( A \) and \( B \). Through algebra, we find that \( A = 1 \) and \( B = -1 \). This allows us to express the integrand as two simpler terms: \( \frac{1}{x-1} - \frac{1}{x+1} \).
Partial fraction decomposition transforms a complex rational function into a simple sum that is easier to integrate.

In the realm of calculus, especially when dealing with improper integrals, determining whether an integral converges or diverges is crucial. An improper integral can have infinite limits or an integrand that becomes unbounded. In such cases, we look at whether the area under the curve approaches a finite value (convergence) or not (divergence).
In our exercise, we evaluated:

• \( \int_{3}^{+\infty} \left( \frac{1}{x-1} - \frac{1}{x+1} \right) dx \)

By integrating, we found:

• \( \ln \left| \frac{x-1}{x+1} \right| \)

Evaluating the limit as \( x \to \infty \), this expression tends towards 0. At the lower limit, \( x = 3 \), it evaluates to another finite value. Thus, the integral converges, resulting in a finite result of \( \ln 2 \).
This systematic approach helps ensure that we're accurately predicting whether the "area" the function describes is bounded.

The concept of limits is fundamental in calculus and underpins many techniques, including evaluating improper integrals. A limit helps us understand what value a function approaches as the input becomes very large or very small.
In our example, we calculated:

• \( \lim_{x \to \infty} \ln \left| \frac{x-1}{x+1} \right| \)

As \( x \) becomes infinitely large, the ratio \( \frac{x-1}{x+1} \) gets closer and closer to 1. The natural logarithm of 1 is 0, indicating that as \( x \to \infty \), the entire expression approaches 0.
By using limits, we determined that the improper integral converges to \( \ln 2 \) because the infinite part of the integral approaches 0, leaving behind a finite result that can be computed at the specified lower limit.