Problem 34 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Set up the Partial-Fraction Decomposition

We express the fraction \( \frac{1}{(x-1)(x+2)} \) as a sum of fractions: \( \frac{A}{x-1} + \frac{B}{x+2} \). The goal is to determine the constants \( A \) and \( B \).

Step 2: Solve for Constants A and B

Multiply both sides of the equation by \((x-1)(x+2)\) to eliminate the denominators, resulting in the equation: \( 1 = A(x+2) + B(x-1) \). Expanding and simplifying gives \( 1 = Ax + 2A + Bx - B \). Combine like terms to get \( 1 = (A+B)x + (2A-B) \).

Step 3: Equate Coefficients

For the equation \( 1 = (A+B)x + (2A-B) \) to hold for all \( x \), the coefficients of corresponding terms must be equal. Set the coefficient of \( x \) to \( 0 \), giving \( A+B=0 \), and set the constant term equal to \( 1 \), giving \( 2A-B=1 \).

Step 4: Solve the System of Equations

From the equations \( A+B=0 \) and \( 2A-B=1 \): Substitute \( B = -A \) into \( 2A - B = 1 \) to get \( 2A - (-A) = 1 \), which simplifies to \( 3A = 1 \). Solve for \( A \) to get \( A = \frac{1}{3} \). Using \( B = -A \), we find \( B = -\frac{1}{3} \).

Step 5: Rewrite the Integrals

With \( A = \frac{1}{3} \) and \( B = -\frac{1}{3} \), the integral becomes: \( \int \left( \frac{1/3}{x-1} - \frac{1/3}{x+2} \right) dx \).

Step 6: Integrate Each Term

Integrate each term separately: \( \int \frac{1/3}{x-1} dx - \int \frac{1/3}{x+2} dx \). These integrals result in: \( \frac{1}{3} \ln|x-1| - \frac{1}{3} \ln|x+2| + C \), where \( C \) is the constant of integration.

Key Concepts

Integration by Parts

Integration by parts is a technique designed to solve integrals where two functions are multiplied together. It originates from the product rule for differentiation and can be immensely useful for complex integrals. The basic formula is:

• \( \int u \, dv = uv - \int v \, du \)

Here, you first identify part of the integrand as \( u \) and another part as \( dv \). After differentiating \( u \) to find \( du \), and integrating \( dv \) to find \( v \), you substitute these into the formula. Remember, the choice of \( u \) and \( dv \) can greatly affect the simplicity of the solution. Pick them in a way that simplifies the subsequent integration process.
Integration by parts is not directly used in partial-fraction decomposition, but understanding it can aid in breaking down complex integrals if needed. It's like having an extra tool in your problem-solving toolkit.

System of Equations

Solving a system of equations involves finding values for variables that satisfy all equations simultaneously. In partial-fraction decomposition, you often derive a system of linear equations to find unknown constants.

• For example, from \( 1 = (A+B)x + (2A-B) \):
• We set the coefficient of \( x \) to zero, giving \( A + B = 0 \).
• The constant term equals one, so \( 2A - B = 1 \).

By solving these equations together, you find \( A \) and \( B \) accurately.
This process relies heavily on your ability to manipulate and solve equations, which is a foundational skill in algebra. Remember to consider each equation's role and use substitution or elimination methods for solutions. System of equations provides a structured way to decode complex algebraic expressions into simpler components.

Natural Logarithms

Natural logarithms are the logarithms to the base \( e \), where \( e \) is a mathematical constant approximately equal to 2.71828. They are usually denoted as \( \ln \). Natural logarithms are particularly useful in integration because of their straightforward derivatives and antiderivatives.

• The derivative of \( \ln|x| \) is \( \frac{1}{x} \), making it a natural choice when dealing with integrals of the form \( \frac{1}{x} \).
• In our context, the integral of \( \frac{1}{x-a} \) is \( \ln|x-a| \) which simplifies calculations.

When working with partial-fraction decomposition, you often encounter terms like \( \int \frac{1/3}{x-1} dx \), which integrate to natural logarithms such as \( \frac{1}{3} \ln|x-1| \).
It's essential to embrace natural logarithms in calculus, recognizing their frequent appearances and benefits they offer in simplifying integration tasks.

Mathematical Constants

Mathematical constants like \( e \) and \( \pi \) are fundamental numbers that appear consistently across various areas of mathematics. In particular, \( e \) is the base of natural logarithms and is key to many calculus concepts.

• \( e \) is approximately 2.71828 and has unique properties such as \( e^0 = 1 \) and it serves as the limit of \( (1 + \frac{1}{n})^n \) as \( n \to \infty \).
• These constants provide the foundation for exponential functions and are crucial in understanding growth patterns, especially in natural processes and finance.

In integration, constants like \( C \) are often added to denote the family of antiderivatives since indefinite integrals can vary by a constant value. This ensures that we have the complete solution set.
Develop a familiarity with these constants as they often simplify complex mathematical concepts, making them a recurring theme in calculus and beyond.