/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Evaluate the integral. $$\int ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 14

Evaluate the integral. $$\int \frac{x^{2}+2 x-1}{x^{3}-x} d x$$

Short Answer

Expert verified
\( \ln \left| \frac{x(x-1)}{x+1} \right| + C \) is the integral.

Step by step solution

01

Simplify the Integrand by Partial Fraction Decomposition

First, we notice that the denominator can be factored. We have \( x^3 - x = x(x-1)(x+1) \). We will use partial fraction decomposition to express our integrand in a simpler form. We write:\[ \frac{x^2 + 2x - 1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}, \]where \(A\), \(B\), and \(C\) are constants to be determined.
02

Setup Equations for Constants

Multiply both sides by \(x(x-1)(x+1)\) to clear the fractions:\[ x^2 + 2x - 1 = A(x-1)(x+1) + B(x)(x+1) + C(x)(x-1). \]Expand the right-hand side:\[ A(x^2 - 1) + B(x^2 + x) + C(x^2 - x). \]This results in:\[ Ax^2 - A + Bx^2 + Bx + Cx^2 - Cx. \]Combine like terms:\[ (A + B + C)x^2 + (B - C)x - A. \]Now, equate coefficients with the left-hand side: - For \(x^2\): \(A + B + C = 1\) - For \(x\): \(B - C = 2\) - Constant: \(-A = -1\).
03

Solve for the Constants

From the third equation, we get:\[ A = 1. \]Substitute \(A = 1\) in the first equation:\[ 1 + B + C = 1 \implies B + C = 0. \]From the second equation:\[ B - C = 2. \]We now have the system of equations:\[ B + C = 0 \]\[ B - C = 2. \]Solving these, we find:\[ B = 1, \ C = -1. \]
04

Write the Partial Fraction Decomposition

Substitute the values of \(A\), \(B\), and \(C\) back into our partial fraction decomposition:\[ \frac{1}{x} + \frac{1}{x-1} - \frac{1}{x+1}. \]
05

Integrate Each Term Separately

Now, we integrate each term from the partial fraction decomposition:- \( \int \frac{1}{x} \, dx = \ln|x| + C_1 \)- \( \int \frac{1}{x-1} \, dx = \ln|x-1| + C_2 \)- \( \int -\frac{1}{x+1} \, dx = -\ln|x+1| + C_3 \)So, the integral becomes:\[ \ln|x| + \ln|x-1| - \ln|x+1| + C. \]
06

Combine Logarithms

Combine the logarithms using properties of logs:\[ \int \frac{x^2+2x-1}{x^3-x} \, dx = \ln \left| \frac{x(x-1)}{x+1} \right| + C. \]

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Key Concepts

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

Integral Calculus
Integral calculus is a core part of calculus that explores how to find the antiderivative, also known as the integral. Here, we evaluate the area under a curve defined by a function. The fundamental theorem of calculus links differentiation and integration, two main operations in calculus. In this exercise, we aim to find the integral of a rational function. A rational function consists of polynomials in both the numerator and the denominator.
  • Notation: Integrals are typically written with a stylized 'S' (鈭�), followed by the function and the differential of the variable (e.g., \( \int f(x) \, dx \)).
  • Goal: The aim is to find the antiderivative, which, when differentiated, will yield the original function.
  • Challenge: Solving integrals can be straightforward, but they become trickier when dealing with complex expressions or functions that do not have straightforward antiderivatives.
In our problem, after simplifying the integrand with partial fraction decomposition, the task becomes more manageable. This method rewrites the expression in a form that is easier to integrate.
Logarithmic Integration
Logarithmic integration is a technique used when the antiderivative of a function results in a logarithm function. This technique is especially useful for rational functions where the numerator is the derivative of the denominator or can be rewritten as one.
In this exercise, after applying partial fraction decomposition, each term in the expression we integrate becomes a simple rational expression of the form \(\frac{1}{x}\), \(\frac{1}{x-1}\), or \(\frac{1}{x+1}\). These types of integrals lead directly to natural logarithm functions when integrated:
  • \(\int \frac{1}{u} \, du = \ln|u| + C\), where \(u\) is a function of \(x\).
By applying this rule, we simplify the integral problem into three separate, easily solvable logarithmic integrations. Finally, we can use the properties of logarithms to combine the results for a succinct final expression.
Coefficient Comparison
Coefficient comparison is a vital algebraic technique used to solve for unknown constants in partial fraction decomposition. This method aligns coefficients of similar powers of \(x\) from both sides of an equation.
  • Purpose: To determine the constants in the decomposed form of a rational expression.
  • Process: Expand both sides of the equation and collect like terms.
  • Compare: Match the coefficients of each power of \(x\) to form system of equations, which are then solved by substitution or elimination.
In our exercise, setting up equations from coefficients allows us to find the values of \(A\), \(B\), and \(C\). Solving these equations provides the values necessary to rewrite the integrand in a decomposed form, enabling easier integration.

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