Problem 30 - Calculus Single Variable

Question

Use the method of partial fractions to calculate the given integral. $$ \int \frac{x^{3}-2 x^{2}-2 x-2}{x^{2}-x} d x $$

Step-by-Step Solution

Answer

\( \frac{x^2}{2} + x + \ln|x| - 2\ln|x-1| + C \) is the integral.

Step 1: Simplify the integrand

Perform polynomial long division to simplify the fraction \( \frac{x^3 - 2x^2 - 2x - 2}{x^2 - x} \). Divide the numerator \( x^3 - 2x^2 - 2x - 2 \) by the denominator \( x^2 - x \). This results in a quotient and a remainder.

Step 2: Express the fraction with the remainder

The polynomial division gives a quotient of \( x + 1 \) with a remainder of \(-x - 1\). Thus, the original fraction can be expressed as:\[\frac{x^3 - 2x^2 - 2x - 2}{x^2 - x} = x + 1 + \frac{-x - 1}{x^2 - x}\]Rewrite this expression to facilitate integration.

Step 3: Decompose the fraction into partial fractions

Decompose the fraction \( \frac{-x - 1}{x^2 - x} \) into partial fractions. Since the denominator is \( x(x - 1) \), we express it as:\[\frac{-x - 1}{x(x - 1)} = \frac{A}{x} + \frac{B}{x - 1}\] Find constants \( A \) and \( B \).

Step 4: Solve for constants A and B

Set up the equation:\[-x - 1 = A(x - 1) + Bx\]Expanding gives \(-x - 1 = Ax - A + Bx\). By comparing coefficients, we find two equations:\( A + B = -1 \) and \(-A = -1\). Solving gives \( A = 1 \) and \( B = -2 \).

Step 5: Integrate each term separately

Now integrate the terms separately:\[\int (x + 1)\, dx + \int \frac{1}{x}\, dx - 2 \int \frac{1}{x-1} \ dx\]Integrate to get:\[\frac{x^2}{2} + x + \, \ln|x| - 2\, \ln|x-1| + C\]where \( C \) is the constant of integration.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used in calculus to break down complex rational expressions into simpler fractions. The idea is to express a single, complicated fraction into a sum of simpler terms that are easier to integrate. In the context of integration, these simpler fractions simplify the integration process.

When decomposing into partial fractions, you usually focus on the denominator. If it's factorable, like in our exercise, you factor the polynomial first. In our case, we had a denominator of \( x(x - 1) \). Then, you express the original fraction as a sum of fractions with unknown coefficients over these factors.

• The step involves expressing the fraction as \( \frac{A}{x} + \frac{B}{x-1} \).
• The next step is finding the values for \( A \) and \( B \).

This involves equating the original fraction to the sum of the decomposed fractions and solving for the unknowns. In our example, this algebraic manipulation helped us find \( A = 1 \) and \( B = -2 \). The separate fractions are now poised for easy integration.

Polynomial Long Division

Polynomial long division is an essential algebraic technique, especially when dealing with the integration of rational functions. It works similarly to numerical long division, and it's used to simplify complex polynomial expressions.

In the given exercise, we used polynomial long division to divide the numerator \( x^3 - 2x^2 - 2x - 2 \) by the denominator \( x^2 - x \). This procedure broke the fraction down into a simpler form that included a quotient and a remainder.

• The quotient in this case was \( x + 1 \).
• The remainder was \( -x - 1 \).

Thus, our original expression \( \frac{x^3 - 2x^2 - 2x - 2}{x^2 - x} \) was expressed as \( x + 1 + \frac{-x - 1}{x^2 - x} \). This step was crucial because it turned a complex polynomial into a manageable expression for further decomposition and integration.

Integration Techniques

Once the expression has been simplified through decomposition into partial fractions, each component can be integrated using basic integration techniques. These techniques leverage the properties of integrals and make use of known integral results.

In our exercise, after decomposition, we ended up with separate components: \( x + 1, \frac{1}{x}, \) and \( \frac{1}{x-1} \). Each of these components is integrated independently.

• The term \( \int (x + 1) \, dx \) results in \( \frac{x^2}{2} + x \).
• The term \( \int \frac{1}{x} \, dx \) yields \( \ln|x| \).
• The term \( -2 \int \frac{1}{x-1} \, dx \) leads to \( -2 \, \ln|x-1| \).

The result is then combined to obtain the final integrated function: \( \frac{x^2}{2} + x + \ln|x| - 2 \ln|x-1| + C \). Constant \( C \) represents an arbitrary constant of integration, a reminder that our integral represents a family of functions.

Rational Functions

Rational functions are expressions that are the ratio of two polynomials. In calculus, they are common targets for integration due to their wide application in many fields.

In our problem, the rational function \( \frac{x^3 - 2x^2 - 2x - 2}{x^2 - x} \) appeared as a complex polynomial division. To effectively integrate rational functions, we must simplify them into simpler pieces, like linear or quadratic terms, often through partial fraction decomposition.

• Rational functions can sometimes be simplified through polynomial long division or direct factoring.
• Understanding the behavior of these functions, especially at the asymptotic and intercept values, can be key in solving calculus problems.

In practice, mastering the techniques of handling rational functions, like decomposition and division, equips students to tackle various complex integrals they may encounter throughout their studies.