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Chapter 5: Problem 37

$$ \text { Evaluate using CAS. } $$ $$ \int \frac{1}{x\left(x^{2}-1\right)} d x $$

Short Answer

Expert verified
\( \int \frac{1}{x(x-1)(x+1)} dx = A \ln |x| + B \ln |x-1| + C \ln |x+1| + C' \)

Step by step solution

01

Factor the Denominator

Rewrite the integral by factoring the denominator. Notice that the denominator can be factored as \( x(x^2 - 1) = x(x - 1)(x + 1) \). The integral becomes: \[ \ \begin{aligned} \ \int \frac{1}{x(x-1)(x+1)} dx \ \ \end{aligned} \]
02

Set Up Partial Fractions

Express the integrand as a sum of partial fractions: \[ \ \begin{aligned} \ \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} \ \ \end{aligned} \]
03

Solve for Constants A, B, and C

Multiply both sides by the common denominator \( x(x-1)(x+1) \) and solve for constants A, B, and C: \[ \ \begin{aligned} \ 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1) \ \ \end{aligned} \] Expand and collect like terms to create a system of equations to solve for A, B, and C.
04

Integrate Each Partial Fraction

Integrate each of the fractions separately: \[ \ \begin{aligned} \ \int \frac{A}{x} dx + \int \frac{B}{x-1} dx + \int \frac{C}{x+1} dx \ \ \end{aligned} \] Recall that the integral of \( \frac{1}{x} \) is \( \ln |x| + C \).
05

Combine Results

Combine the resulting integrals to find the final answer: \[ \ \begin{aligned} \ \int \frac{1}{x(x-1)(x+1)} dx = A \ln |x| + B \ln |x-1| + C \ln |x+1| + C' \ \ \end{aligned} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integration
Definite integration involves finding the integral of a function over a specific interval. It provides the accumulated quantity between two points, say \( a \) and \( b \). The result of a definite integral is a number, unlike indefinite integration which results in a family of functions. To compute a definite integral, you evaluate the antiderivative at the upper limit and subtract the value at the lower limit:
  • Notation: \( \int_{a}^{b} f(x) \, dx \)
  • Example: \( \int_{0}^{2} x^2 \, dx = [\frac{x^3}{3}]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \)
Definite integration has applications like computing areas under curves, accumulated quantities, and solving problems in physics and engineering.
Indefinite Integration
Indefinite integration, also known as antiderivative, represents a family of functions whose derivative is the integrand. It is typically written with a constant of integration, \( C \), since the derivative of any constant is zero. The result of an indefinite integral is a function plus this constant:
  • Notation: \( \int f(x) \, dx = F(x) + C \)
  • Examples: \( \int x^2 \, dx = \frac{x^3}{3} + C \) and \( \int e^x \, dx = e^x + C \)
Indefinite integrals are fundamental in solving differential equations, finding original functions, and understanding mathematical relationships.
Integration Techniques
Integration techniques are strategies to find integrals that are not straightforward. Here are a few common techniques:
  • Substitution: Simplifies the integral by substituting part of the integrand with a new variable. For example, \( \int e^{2x} \, dx \) can be solved by substituting \( u = 2x \).
  • Integration by Parts: Based on the product rule of differentiation. For example, \( \int x e^x \, dx \) can be solved as \( uv - \int v \, du \).
  • Partial Fractions: Decomposes a rational function into simpler fractions that are easier to integrate. This involves expressing the integrand as a sum of simpler rational expressions. For instance, in your exercise, the given fraction is decomposed into partial fractions for easier integration.
Learning these techniques expands your toolkit for addressing a variety of integrals in calculus, making complex problems simpler and more manageable.

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