Problem 41 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \frac{x-1}{x(x-2)} d x=\) (A) \(\frac{1}{2} \ln |x|+\ln |x-2|+C\) (B) \(\frac{1}{2} \ln \left|\frac{x-2}{x}\right|+C\) (C) \(\frac{1}{2} \ln |x(x-2)|+C\) (D) \(\ln \left|\frac{x}{x-2}\right|+C\)

Step-by-Step Solution

Answer

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

Step 1: Set up Partial Fraction Decomposition

The function \( \frac{x-1}{x(x-2)} \) can be expressed as a sum of partial fractions: \( \frac{A}{x} + \frac{B}{x-2} \). Set up the equation: \[ \frac{x-1}{x(x-2)} = \frac{A}{x} + \frac{B}{x-2} \] Multiply through by the denominator \( x(x-2) \) to get: \[ x-1 = A(x-2) + Bx \].

Step 2: Solve for Coefficients A and B

Expand the equation: \[ x-1 = Ax - 2A + Bx \] Rearrange terms: \[ x - 1 = (A + B)x - 2A \] This implies two equations: 1. \( A + B = 1 \) 2. \( -2A = -1 \) Solve for A: \( A = \frac{1}{2} \). Substitute \( A \) into the first equation: \( \frac{1}{2} + B = 1 \), which gives \( B = \frac{1}{2} \).

Step 3: Write the Integral with Partial Fractions

Substitute \( A \) and \( B \) back into the partial fractions: \[ \frac{1}{2} \int \frac{1}{x} \, dx + \frac{1}{2} \int \frac{1}{x-2} \, dx \].

Step 4: Integrate Each Term Individually

Integrate each partial fraction: 1. \( \int \frac{1}{x} \, dx = \ln |x| \) 2. \( \int \frac{1}{x-2} \, dx = \ln |x-2| \) Therefore, \[ \frac{1}{2} \ln |x| + \frac{1}{2} \ln |x-2| + C \].

Step 5: Combine the Results into a Single Expression

Apply the properties of logarithms: \[ \frac{1}{2} \ln |x| + \frac{1}{2} \ln |x-2| = \frac{1}{2} \ln |x(x-2)| \]. Therefore, the solution to the integral is: \[ \frac{1}{2} \ln |x(x-2)| + C \].

Key Concepts

Partial fraction decomposition

In calculus, partial fraction decomposition is a technique used to simplify complex rational expressions into simpler fractions, making integration easier. When you encounter a fraction with a polynomial in the numerator and a polynomial in the denominator, it can often be broken down into a sum of simpler fractions, called partial fractions.

For example, in the exercise given, the expression \( \frac{x-1}{x(x-2)} \) is decomposed into the form \( \frac{A}{x} + \frac{B}{x-2} \). By doing this, we express the original complex fraction as a sum of fractions with linear denominators.

• Set Up: Decomposition begins by equating the complex fraction to a sum of its partial fractions. In our exercise, it's \( \frac{A}{x} + \frac{B}{x-2} \).
• Find Coefficients: Multiply through by the original denominator to eliminate the fractions, resulting in an equation that can be solved for \( A \) and \( B \). In the solution, solving the equations \( A + B = 1 \) and \(-2A = -1\) gives us \( A = \frac{1}{2} \) and \( B = \frac{1}{2} \).
• Back-Substitution: Substitute the determined values of \( A \) and \( B \) back into the partial fractions. This simplifies the original integral problem, paving the way for easier integration of each part separately.

Understanding partial fraction decomposition is crucial in handling rational expressions and eases the process of integration.

Logarithmic integration

Logarithmic integration is a method used to integrate expressions of the form \( \int \frac{1}{x} \, dx \), resulting in a logarithmic function. This method is particularly useful when dealing with partial fractions like those seen in our example.

The integral of expressions involving \( \frac{1}{x} \) and \( \frac{1}{x-a} \) directly relate to the natural logarithm function.

• Integration Basics: When you integrate \( \int \frac{1}{x} \, dx \), you obtain \( \ln |x| \). This integration principle is applied to each term of the decomposed fractions.
• Apply to Partial Fractions: Integrate each partial fraction individually, such as \( \frac{1}{2} \int \frac{1}{x} \, dx = \frac{1}{2} \ln |x| \) and \( \frac{1}{2} \int \frac{1}{x-2} \, dx = \frac{1}{2} \ln |x-2| \). This results in a log term for each part of the decomposed fractions.
• Combining Logs: You can often combine these log terms using properties of logarithms. In the given problem, \( \frac{1}{2} \ln |x| + \frac{1}{2} \ln |x-2| \) can be combined as \( \frac{1}{2} \ln |x(x-2)| \).

Logarithmic integration simplifies the handling of certain rational expressions, providing a straightforward path to solve integrals.

Indefinite integrals

In the realm of calculus, an indefinite integral represents a function that describes how an area under a curve accumulates. Unlike definite integrals that result in a numerical value, indefinite integrals focus on finding a family of functions, which include a constant of integration \( C \).

Indefinite integrals are expressed as:

• Definition: The indefinite integral of a function \( f(x) \), symbolized as \( \int f(x) \, dx \), yields a function \( F(x) + C \) such that \( F'(x) = f(x) \).
• Constant of Integration: The \( C \) represents an arbitrary constant. It accounts for all possible vertical shifts of the antiderivative since the derivative of a constant is zero.
• Integrating Rational Functions: When dealing with an expression such as \( \frac{x-1}{x(x-2)} \), breaking it into partial fractions simplifies the integration process. Each partial fraction, when integrated, contributes to the indefinite integral.

In the given problem, the expression \( \frac{x-1}{x(x-2)} \) is integrated, yielding \( \frac{1}{2} \ln |x(x-2)| + C \). The constant \( C \) is crucial because it signifies that there are infinitely many solutions, each differing by a constant amount.