Problem 16 - Calculus for Biology and Medicine

Question

In Problems , use partial-fraction decomposition to evaluate the integrals.$$ \int \frac{1}{(x-1)(x+2)} d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{3} \ln |x-1| - \frac{1}{3} \ln |x+2| + C \).

Step 1: Setup the Partial Fraction Decomposition

First, express the integrand \( \frac{1}{(x-1)(x+2)} \) as a sum of partial fractions. Assume \( \frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \).

Step 2: Clear the Denominators

Multiply through by \((x-1)(x+2)\) to eliminate the denominators. You get \(1 = A(x+2) + B(x-1)\).

Step 3: Expand and Simplify

Expand the right side: \(1 = Ax + 2A + Bx - B\). Combine like terms: \(1 = (A + B)x + (2A - B)\).

Step 4: Set Up Equations and Solve for Coefficients

Equate coefficients from both sides. We get two equations: \(A + B = 0\) and \(2A - B = 1\). Solve these equations simultaneously to get \(A = \frac{1}{3}\) and \(B = -\frac{1}{3}\).

Step 5: Integrate Both Terms

Rewrite the original integrand using the partial fractions: \( \frac{1}{3(x-1)} - \frac{1}{3(x+2)} \). Integrate each term separately: \(\int \frac{1}{3(x-1)} \, dx - \int \frac{1}{3(x+2)} \, dx\).

Step 6: Calculate the Integral

The integrals are: \( \frac{1}{3} \ln |x-1| - \frac{1}{3} \ln |x+2| + C \) where \(C\) is the integration constant.

Key Concepts

Integration Techniques

When tackling integrals that involve complex expressions, specific integration techniques such as partial fraction decomposition come in handy. These techniques are methods used to break down integrals into simpler, more recognizable, and often easier-to-integrate pieces.

• Partial Fraction Decomposition: This method is particularly useful for integrating rational functions, where the numerator is of a lower degree than the denominator. It involves expressing a rational function as a sum of simpler fractions that can be easily integrated.
• Decomposition Steps: Firstly, represent the complex rational expression as a sum of partial fractions, each having simpler denominators. Then, solve for unknown coefficients by setting up equations through equating coefficients from both sides of the equation.

By applying these integration techniques, especially partial fraction decomposition, we simplify the original integral, ultimately making the process of integration straightforward. Techniques like these are valuable in calculus when dealing with otherwise complicated integral expressions.

Calculus Integrals

In calculus, integrals are fundamental concepts that represent the area under a curve or the accumulation of quantities. Calculus integrals can be classified into two main types: definite and indefinite integrals.

• Indefinite Integrals: Represent a family of functions with an arbitrary constant added, often denoted as 'C'. For instance, when integrating partial fractions, the result is an indefinite integral since it includes an integration constant.
• Definite Integrals: Compute the net area between the curve and the x-axis over a specified interval. Although the exercise at hand focuses on indefinite integrals, it’s important to understand the role of definite integrals in calculating exact quantities.
• Logarithmic Involvement: Often in partial fraction decomposition, especially when the decomposed terms are linear, the result involves logarithmic functions, as seen in the solution with \( \ln |x-1| \) and \( \ln |x+2| \).

Understanding these integral types helps in recognizing when and how to apply techniques such as partial fraction decomposition effectively to solve real-world calculus problems.

Rational Functions

Rational functions are ratios of two polynomials. Understanding them is crucial, as they regularly appear in calculus problems requiring integration, including the partial fraction decomposition process.

• Understanding Rational Functions: A rational function is expressed as \( \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). These functions can describe various natural phenomena and are widely used in calculus.
• Decomposition Importance: The key advantage of decomposing a complex rational function into partial fractions is simplifying the integration process. It breaks down complicated expressions into a sum of simpler terms, each of which is easier to integrate.
• Application in Integration: By applying partial fractions, we reduce the degree of the polynomial in the denominator, allowing for straightforward integration with basic calculus methods. This results in expressions where logarithmic or simple polynomial integrating techniques suffice.

Recognizing and manipulating rational functions through decomposition is essential, not only for simplifying integrals but also for developing a deeper understanding of calculus and its practical applications.