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Evaluate the following integrals. $$\int \frac{4 x-2}{x^{3}-x} d x$$

Short Answer

Expert verified
The values of A, B, and C are found by solving the equation obtained after multiplying both sides of the decomposition by the denominator, which enables us to eliminate the denominators on the right side of the equation. This step involves setting the values of x that make each term zero and solving for the constants.

Step by step solution

01

Factor the denominator

Factor the denominator by taking a common factor out. In this case, we can take x out since it is common to both terms in the denominator: $$x^3 - x = x(x^2 - 1)$$ Now, we can further factor the term \((x^2 - 1)\) as it is a difference of squares: $$x^2 - 1 = (x - 1)(x + 1)$$ So, after factoring, the integral becomes: $$\int \frac{4x - 2}{x(x - 1)(x + 1)} dx$$
02

Perform partial fraction decomposition

Using partial fraction decomposition, we can rewrite the integral as the sum of three simpler fractions: $$\frac{4x - 2}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$$ To find the values of A, B, and C, we can multiply both sides by the denominator \(x(x - 1)(x + 1)\) to eliminate the denominators on the right side of the equation: $$(x(x-1)(x+1))\left( \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}\right) = 4x - 2$$ Which simplifies to: $$A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1) = 4x - 2$$ Now, we can find the values of A, B, and C by solving this equation. We can do this by setting the values of x that make each term zero and solving for the constants. For A, let \(x=0\): $$A(-1)(1)=0-2$$ $$A = 2$$ For B, let \(x=1\): $$B(1)(1+1)=0-2$$ $$2B=-2$$ $$B=-1$$ For C, let \(x=-1\): $$C(-1)(-1-1)=0-2$$ $$2C=-2$$ $$C=-1$$ Now, rewrite the integral with the values of A, B, and C: $$\int \frac{4x - 2}{x(x - 1)(x + 1)} dx = \int \left( \frac{2}{x} - \frac{1}{x - 1} - \frac{1}{x + 1}\right)dx$$
03

Integrate each term of the decomposed fraction

Now that we have the decomposed fractions, integrate each term: $$\int \frac{2}{x} dx - \int\frac{1}{x - 1} dx - \int\frac{1}{x + 1} dx$$ Each term can be easily integrated using the power rule: $$\int \frac{2}{x}dx = 2 \ln|x| + C_1$$ $$\int\frac{1}{x - 1}dx = \ln|x - 1| + C_2$$ $$\int\frac{1}{x + 1}dx = \ln|x + 1| + C_3$$
04

Combine results and simplify

Combine the results: $$2\ln|x| - \ln|x - 1| - \ln|x + 1| + C$$ Where, \(C = C_1 + C_2 + C_3\) is an arbitrary constant. Therefore, the final result is: $$\int \frac{4x-2}{x^3-x} dx = 2\ln|x| - \ln|x - 1| - \ln|x + 1| + C$$

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Key Concepts

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

Partial Fraction Decomposition
Partial fraction decomposition is a technique used to simplify the integration of complex rational functions. This method is particularly useful when the integrand involves a rational function, which is a fraction where both the numerator and the denominator are polynomials.

In the given exercise, we start with the rational function \( \frac{4x-2}{x^3-x} \). To decompose it, we first factor the denominator, \( x^3-x \), into \( x(x-1)(x+1) \). This allows us to rewrite the original fraction as a sum of simpler fractions, each with a denominator that is one of the factors. The decomposition takes the form:
\[\frac{4x - 2}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}\]

We solve for the constants \( A, B, \) and \( C \) by clearing the denominators and equating coefficients, resulting in simpler fractions that are much easier to integrate individually. This strategic breakdown into partial fractions transforms a difficult integration problem into multiple manageable ones.

Partial fraction decomposition is essential in solving integrals involving rational fractions, especially when the degree of the numerator is less than that of the denominator.
Integration Techniques
Integration techniques refer to various methods used to evaluate integrals, each applicable in different scenarios to simplify or solve an integral. The exercise uses several integration techniques to solve the rational function.

- **Factoring**: The first step involves factoring the denominator. Factoring simplifies the expression and allows us to apply partial fraction decomposition. By breaking down \( x^3 - x \) into \( x(x-1)(x+1) \), we set the stage for further simplification.

- **Partial Fraction Decomposition**: This technique simplifies the rational function into simpler fractions, each of which can be integrated using basic integral rules.

- **Basic Integration Rules**: Once the function is decomposed, we apply integration rules for logarithmic functions to solve each term. These involve direct integration, making it straightforward to achieve the final solution.

Each technique is crucial for solving integrals efficiently, and understanding when and how to apply them is foundationally important in calculus.
Logarithmic Integration
Logarithmic integration is a method where integrals involve logarithmic functions or result in logarithmic functions after evaluation. This becomes particularly straightforward when dealing with terms of the form \( \frac{1}{x} \).

In our exercise, after decomposing the rational function and setting it up for integration, we integrate each part separately:
\[\int \frac{2}{x} dx - \int \frac{1}{x - 1} dx - \int \frac{1}{x + 1} dx\]

Each of these integrals results in a logarithmic function:
  • \( \int \frac{2}{x} dx = 2 \ln|x| + C_1 \)
  • \( \int \frac{1}{x - 1} dx = \ln|x - 1| + C_2 \)
  • \( \int \frac{1}{x + 1} dx = \ln|x + 1| + C_3 \)

Logarithmic integration often results in neat expressions as it uses the natural logarithm function, where the constant \( C \) represents the constant of integration. This technique is essential for integrating rational functions where the derivative of the denominator is in the numerator, or the integral of each fraction form results in simple logarithms.
Rational Functions
Rational functions are fractions where both the numerator and the denominator are polynomials. They are a central focus in integral calculus due to their ubiquity and the complexity they sometimes exhibit when integrating.

In the problem we solved, the function \( \frac{4x-2}{x^3-x} \) is a classic example of a rational function with a degree greater in the denominator. Such functions require special techniques to integrate effectively.

- **Factoring**: Simplifying the denominator by factoring is a fundamental step.- **Partial Fractions**: Decomposing the function into simpler fractions makes integration possible.

Rational functions can range from simple to highly complex, depending on the degree and factorization of their polynomials. Understanding how to break them down using partial fraction decomposition leads to successful integration by allowing each term to be manageable through direct integration rules.

Learning to manage rational functions through these methods is a powerful skill in calculus that applies across various mathematical, physical, and engineering applications.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. The Trapezoid Rule is exact when used to approximate the definite integral of a linear function. b. If the number of subintervals used in the Midpoint Rule is increased by a factor of \(3,\) the error is expected to decrease by a factor of 8. c. If the number of subintervals used in the Trapezoid Rule is increased by a factor of \(4,\) the error is expected to decrease by a factor of 16.

The Eiffel Tower property Let \(R\) be the region between the curves \(y=e^{-\alpha x}\) and \(y=-e^{-\alpha x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c>0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves; see the Guided Project The exponential Eiffel Tower.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\).

Show that \(L=\lim _{n \rightarrow \infty}\left(\frac{1}{n} \ln n !-\ln n\right)=-1\) in the following steps. a. Note that \(n !=n(n-1)(n-2) \cdots 1\) and use \(\ln (a b)=\ln a+\ln b\) to show that $$ \begin{aligned} L &=\lim _{n \rightarrow \infty}\left(\left(\frac{1}{n} \sum_{k=1}^{n} \ln k\right)-\ln n\right) \\ &=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \frac{k}{n} \end{aligned} $$ b. Identify the limit of this sum as a Riemann sum for \(\int_{0}^{1} \ln x d x\) Integrate this improper integral by parts and reach the desired conclusion.

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=-\ln x\) and the \(x\) -axis on the interval (0,1] is revolved about the \(x\) -axis.

Approximate the following integrals using Simpson's Rule. Experiment with values of \(n\) to ensure that the error is less than \(10^{-3}\). \(\int_{0}^{2 \pi} \frac{d x}{(5+3 \sin x)^{2}}=\frac{5 \pi}{32}\)

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