Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the given integral. $$ \int \frac{x}{x^{2}-5 x+6} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(\ln \left| \frac{(x-2)^2}{x-3} \right| + C\).

Step 1: Factor the denominator

First, we need to factor the quadratic expression in the denominator: \(x^2 - 5x + 6\). This can be factored into \((x - 2)(x - 3)\). Now, our integral looks like \(\int \frac{x}{(x - 2)(x - 3)} \, dx\).

Step 2: Set up partial fraction decomposition

Now, express the integrand as a sum of partial fractions: \(\frac{x}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}\).

Step 3: Solve for coefficients A and B

Multiply both sides by \((x - 2)(x - 3)\) to clear the denominators:\[ x = A(x - 3) + B(x - 2) \]Expand and collect like terms:\[ x = Ax - 3A + Bx - 2B \]Combine terms to get:\[ x = (A + B)x - (3A + 2B) \]From this, we equate coefficients to solve for \(A\) and \(B\):1. \( A + B = 1 \)2. \( -3A - 2B = 0 \)From the second equation, solve for one variable in terms of the other, then substitute back into the first equation to find that \(A = 2\) and \(B = -1\).

Step 4: Integrate each partial fraction

The integrand is now \(\frac{2}{x-2} - \frac{1}{x-3}\). Integrate each term separately:\[ \int \left(\frac{2}{x-2} - \frac{1}{x-3}\right) \, dx = 2\ln|x-2| - \ln|x-3| + C \]

Step 5: Simplify the integral expression

Combine the logarithmic expressions:\[ \int \frac{x}{x^2 - 5x + 6} \, dx = \ln \left| \frac{(x-2)^2}{x-3} \right| + C \]

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in calculus to break down complex rational expressions into simpler fractions. This makes them easier to integrate or differentiate.
To use this method, you first factor the denominator of the rational function. Once factored, rewrite the rational function as a sum of simpler fractions, each having one of the factors as its denominator.

• For example, in our exercise, the expression \( \frac{x}{x^2 - 5x + 6} \) was initially given.
• The denominator factors into \((x-2)(x-3)\).
• Using partial fraction decomposition, it is rewritten as \( \frac{A}{x-2} + \frac{B}{x-3} \).

In this representation, \(A\) and \(B\) are constants that need to be determined. We clear the denominators by multiplying through by the factored denominator and then solve for \(A\) and \(B\) by equating the coefficients of the expressions on both sides of the equation. This process simplifies the original expression into partial fractions, which are much easier to integrate, leading us into our next concept.

Integral Calculus

Integral calculus is the branch of mathematics concerning the accumulation of quantities and the areas under curves.
In our exercise, we are tasked with integrating a rational function in partial fraction form. The transformation of the original complex expression into partial fractions simplifies the integration process.

• We need to integrate terms like \( \frac{A}{x-2} \) and \( \frac{B}{x-3} \).
• These terms fit into a standard integral form, allowing direct application of integration rules.

Integrating these simpler functions individually helps us find the antiderivative of the entire expression. Integral calculus thus offers tools and techniques to deal with complex functions by simplifying them into forms that are straightforward to integrate. The results of the integration process provide functions that represent the area under the original function's graph over specified intervals.

Logarithmic Integration

Logarithmic integration is a specific technique for integrating functions, typically involving rational expressions where the denominator is a linear factor. When you integrate expressions like \( \frac{1}{x-a} \), you end up using logarithms. In our problem, after partial fraction decomposition, we had two terms: \( \frac{2}{x-2} \) and \( \frac{-1}{x-3} \).

• Integrating \( \frac{1}{x-2} \) gives \( \ln |x-2| \).
• When integrated, \( \frac{-1}{x-3} \) gives \( -\ln |x-3| \).

These integrals can be combined using the properties of logarithms to simplify the expression further. So, the integration of the function leads to a logarithmic expression that includes terms of natural logarithms because of the linear terms in the denominator. This process helps reveal how closely tied integration and logarithmic rules are, specifically when addressing linear denominators within rational expressions.