91Ó°ÊÓ

Partial fraction decomposition is a technique used to simplify complex rational expressions into simpler, easier-to-integrate fractions. When dealing with integrals like \(\int \frac{1}{t^2-9} \, dt\), recognizing perfect square differences like \(t^2 - 9\) as factorizable terms, i.e., \((t+3)(t-3)\), opens up the method for decomposition.

The goal is to express the fraction \(\frac{1}{t^2-9}\) as a sum of simpler fractions, such as \(\frac{A}{t+3} + \frac{B}{t-3}\). This simplifies the integral into parts that are more straightforward to work with.

• Factorize the denominator of the rational function.
• Set up an equation \(\frac{1}{t^2-9} = \frac{A}{t+3} + \frac{B}{t-3}\) to find constants \(A\) and \(B\).
• Eliminate denominators by multiplying through by \((t+3)(t-3)\) to get a polynomial equation.
• Choose strategic values for \(t\) (often the roots \(t = 3\) and \(t = -3\)), to solve for \(A\) and \(B\) easily.

This approach is particularly helpful for integrating rational functions because it breaks a complex integral into manageable chunks.

Once a function is decomposed into partial fractions, the integration process often involves solving integrals of the form \(\int \frac{1}{x-a} \, dx\), which are simply logarithmic integrals.

In the step-by-step solution, two separate integrals arise from our decomposition process:

• \(-\frac{1}{6} \int \frac{1}{t+3} \, dt\)
• \(\frac{1}{6} \int \frac{1}{t-3} \, dt\)

Such integrals, that involve \(1/(t+c)\), can be directly integrated using the natural logarithm function.

The result of each integral is:

• \(-\frac{1}{6} \ln|t+3|\)
• \(\frac{1}{6} \ln|t-3|\)

Combining these gives the integrated result: \(\frac{1}{6} \ln\left|\frac{t-3}{t+3}\right| + C\).

Remember: Always use the absolute value inside the logarithm to account for any negative values of \(t\) that may arise. This ensures the result remains real and defined across potential inputs.

Understanding the difference between definite and indefinite integrals is crucial in calculus. An indefinite integral aims to find the antiderivative of a function. It does not evaluate the integral over a specific interval but instead results in a function that includes a constant \(C\). For example, the solution provided:

\(\int \frac{1}{t^2 - 9} dt = \frac{1}{6} \ln\left|\frac{t-3}{t+3}\right| + C\) is an indefinite integral.

• Indefinite integrals yield families of functions that differ by a constant \(C\).
• It is denoted without limits on the integral sign.
• The process involves finding antiderivatives and adding a constant \(C\) at the end.

For definite integrals, you compute the net area under a curve between two specified points. This involves evaluating the resulting antiderivative at these points and subtracting them, with no constant of integration \(C\). In the context of indefinite integrals like the one solved, the constant \(C\) represents an unknown constant due to lack of boundary conditions. When limits are given, we use the fundamental theorem of calculus to compute exact values.