23-64. Integration Evaluate the following integrals.
29. ∫₋₁² [(5x) / (x² - x - 6)] dx

Verified step by step guidance

1. First, factor the quadratic expression in the denominator: \(x^{2} - x - 6\). Find two numbers that multiply to \(-6\) and add to \(-1\) to factor it as \((x - 3)(x + 2)\).
2. Rewrite the integral using the factored form: \(\int_{-1}^{2} \frac{5x}{(x - 3)(x + 2)} \, dx\).
3. Set up the partial fraction decomposition for the integrand: \(\frac{5x}{(x - 3)(x + 2)} = \frac{A}{x - 3} + \frac{B}{x + 2}\), where \(A\) and \(B\) are constants to be determined.
4. Multiply both sides by \((x - 3)(x + 2)\) to clear denominators and solve for \(A\) and \(B\) by equating coefficients or substituting convenient values of \(x\).
5. Once \(A\) and \(B\) are found, rewrite the integral as the sum of two simpler integrals: \(\int_{-1}^{2} \frac{A}{x - 3} \, dx + \int_{-1}^{2} \frac{B}{x + 2} \, dx\), and then integrate each term using the natural logarithm function.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down a complex rational function into simpler fractions that are easier to integrate. It involves factoring the denominator and expressing the integrand as a sum of simpler rational expressions. This method is essential when integrating rational functions where the degree of the numerator is less than the degree of the denominator.

Integration of Rational Functions

Integrating rational functions often requires rewriting the integrand into a form that matches standard integral formulas. After partial fraction decomposition, each simpler fraction can be integrated using basic rules, such as integrating 1/(x - a) to get a natural logarithm. Understanding these standard integrals is crucial for solving the problem.

Definite Integration and Limits

Definite integration involves evaluating the integral between two specific limits, which gives the net area under the curve. After finding the antiderivative, you substitute the upper and lower limits and compute the difference. Proper handling of limits ensures the correct evaluation of the integral's value.