Problem 14 - Calculus - AP Edition

Question

Evaluate the integral. $$\int \frac{d x}{x\left(x^{2}-1\right)}$$

Step-by-Step Solution

Answer

\( \int \frac{d x}{x(x^{2}-1)} = -\ln|x| + \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C \)

Step 1: Simplify the integrand

The integral \( \int \frac{d x}{x(x^{2}-1)} \) has a rational function integrand with a quadratic polynomial. To simplify, use partial fraction decomposition to express the integrand as a sum of simpler fractions.

Step 2: Partial Fraction Decomposition

Let \( \frac{1}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} \). Multiply through by \( x(x^2-1) \) to clear the denominators: \( 1 = A(x^2 - 1) + Bx(x+1) + Cx(x-1) \).

Step 3: Solve for coefficients

Equate coefficients of powers of \(x\) from both sides: \( x^2 : A = 0 \), \( x : B + C = 0 \), and \( 1 : -A = 1 \). Solving gives \( A = -1 \), \( B = 1/2 \), and \( C = -1/2 \).

Step 4: Rewrite the integrand

The integrand becomes \( \int \left( \frac{-1}{x} + \frac{1/2}{x-1} + \frac{-1/2}{x+1} \right) d x \).

Step 5: Integrate individually

Integrate term by term: \( \int \frac{-1}{x} \, dx = -\ln|x| + C_1 \), \( \int \frac{1/2}{x-1} \, dx = \frac{1}{2}\ln|x-1| + C_2 \), \( \int \frac{-1/2}{x+1} \, dx = -\frac{1}{2}\ln|x+1| + C_3 \).

Step 6: Combine the results

Combine the terms and constants: \( \int \frac{d x}{x(x^2-1)} = -\ln|x| + \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C \), where \( C = C_1 + C_2 + C_3 \) is the constant of integration.

Key Concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a valuable technique in calculus used to break down complex rational expressions into simpler, more manageable parts. This method is particularly useful when integrating rational functions, which are fractions where both the numerator and the denominator are polynomials. By simplifying a fraction into a series of simpler fractions, the integration process becomes much easier.

For the exercise provided, we decompose the expression into:

• \( \frac{A}{x} \)
• \( \frac{B}{x-1} \)
• \( \frac{C}{x+1} \)

After setting up this equation, multiply throughout by the original denominator to eliminate fractions, facilitating solving for coefficients \( A \), \( B \), and \( C \). By equating terms with similar powers of \( x \), values for these coefficients can be found, resulting in a much simpler form for integration.

The simplicity achieved through partial fraction decomposition allows you to address each part separately, greatly simplifying the integration process.

Rational Function

A Rational Function is one of the fundamental concepts in calculus and algebra, representing any function defined by the ratio of two polynomials. These functions can often appear complex due to their polynomials' nature in both the numerator and the denominator.

In the context of calculus, understanding rational functions is crucial when learning to integrate. The goal is to simplify these functions wherever possible to facilitate easier computation.

Consider the given function \( \frac{1}{x(x^2-1)} \), where the denominator is a product of linear and quadratic polynomials. Such forms lay the groundwork for applying techniques like partial fraction decomposition.

Recognizing and simplifying rational functions is fundamental in calculus exercises because it sets the stage for further operations like differentiation and integration, making arduous calculations manageable.

Integral Calculation

Integral Calculation involves determining the antiderivative of a function, a core concept within the study of calculus. It is essential for resolving areas under curves, among other applications.

Once the function is decomposed using partial fractions, each resulting part can be individually integrated. This segmentation is key in breaking down complex integrals into simpler components that can be solved with standard integration techniques.

For the exercise at hand, the integral is divided into three simpler parts:

• \( \int \frac{-1}{x} \, dx \)
• \( \int \frac{1/2}{x-1} \, dx \)
• \( \int \frac{-1/2}{x+1} \, dx \)

Each of these can be integrated using the natural logarithm rule \( \int \frac{1}{x} \, dx = \ln|x| + C \). Once completed, these terms can be combined to provide the complete solution to the original integral, adding a constant \( C \) to account for the indefinite integral.

Mastering integral calculation through decomposed functions not only aids in solving specific calculus problems but also builds a foundation for tackling more complex mathematical and real-world problems.