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Chapter 8: Problem 33

In Exercises \(33-38,\) perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral. $$\int \frac{2 x^{3}-2 x^{2}+1}{x^{2}-x} d x$$

Short Answer

Expert verified
The integral is \(x^2 - \ln|x| + \ln|x-1| + C\).

Step by step solution

01

Perform Long Division

We start by dividing the polynomial in the numerator by the polynomial in the denominator. Begin with the division: 1. Divide the first term of the numerator, \(2x^3\), by the first term of the denominator, \(x^2\), to get \(2x\).2. Multiply the entire denominator \(x^2 - x\) by \(2x\) to get \(2x^3 - 2x^2\).3. Subtract this from the original numerator: \((2x^3 - 2x^2 + 1) - (2x^3 - 2x^2) = 1\).4. Since \(1\) is the remainder, the quotient is \(2x\) with a remainder of \(1\).Thus, we can rewrite the integrand as \(2x + \frac{1}{x^2 - x}\).
02

Decompose the Proper Fraction

Now, decompose the proper fraction \(\frac{1}{x^2 - x}\) into partial fractions:1. Factor the denominator: \(x^2 - x = x(x-1)\).2. Set up the partial fraction decomposition: \(\frac{1}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1}\).3. Multiply through by the common denominator \(x(x-1)\) to clear the fractions: \[1 = A(x - 1) + Bx\].4. Expand and combine like terms: \(1 = Ax - A + Bx\).5. Equate coefficients to solve for \(A\) and \(B\): - Constant term: \(-A = 1\), so \(A = -1\). - Coefficient of \(x\): \(A + B = 0\), so \(-1 + B = 0\) gives \(B = 1\).The partial fraction decomposition is thus \(-\frac{1}{x} + \frac{1}{x-1}\).
03

Integrate Term-by-Term

Now we integrate term-by-term using the decomposition from Steps 1 and 2:1. Integrate \(2x\): \[\int 2x \, dx = x^2 + C_1\].2. Integrate \(-\frac{1}{x}\): \[\int -\frac{1}{x} \, dx = -\ln|x| + C_2\].3. Integrate \(\frac{1}{x-1}\): \[\int \frac{1}{x-1} \, dx = \ln|x-1| + C_3\].Combine all pieces together: \[x^2 - \ln|x| + \ln|x-1| + C\], where \(C\) is the constant of integration.

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Key Concepts

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

Long Division in Polynomials
Long division for polynomials is a technique that helps us rewrite a complex fraction into something more manageable. Similar to numerical long division, polynomial long division divides one polynomial by another, resulting in a quotient and a remainder. This is crucial when the degree of the polynomial in the numerator is higher than that in the denominator, as seen in the given integral.
**Steps to Perform Long Division:**
  • Divide the first term of the numerator by the first term of the denominator. This gives the first term of the quotient.
  • Multiply the whole divisor by the first term of the quotient and subtract from the original numerator polynomial to get a new polynomial.
  • Repeat these steps with the resulting polynomial until the degree of the new polynomial is less than the degree of the divisor.
For our exercise, the numerator was divided by the denominator, yielding a quotient of \(2x\) and a remainder of \(1\). By rewriting \(\frac{2x^3 - 2x^2 + 1}{x^2 - x}\) as \(2x + \frac{1}{x^2 - x}\), we simplify the task of integration.
Partial Fraction Decomposition
Partial fraction decomposition is a process used to break down complex fractions into simpler components that are easier to work with, particularly during integration. When a polynomial fraction has a factorable denominator, it can often be expressed as a sum of fractions with simpler denominators.
**How to Decompose into Partial Fractions:**
  • Factor the denominator of the fraction into simpler terms, such as linear factors.
  • Create a partial fraction for each factor, assigning unknown coefficients for numerators.
  • Multiply through by the original denominator to eliminate fractions, leading to a polynomial equation.
  • Solve this equation by comparing coefficients, determining the values of the unknowns.
In this instance, the factorization \(x^2-x = x(x-1)\) was used. The corresponding partial fraction decomposition, \(\frac{1}{x(x-1)} = -\frac{1}{x} + \frac{1}{x-1}\), provides simpler forms ready for integration.
Integration Techniques
The integration of polynomial and rational functions often requires breaking them down into simpler parts. Once a function is decomposed into simpler terms through long division and partial fractions, integrating each term separately becomes straightforward using basic integration rules.
**Key Techniques for Integration:**
  • Recognize standard forms, such as \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for a polynomial, and apply appropriately.
  • For rational functions, apply decomposition results allowing for the integration of terms like \(\int \frac{1}{x} \, dx = \ln|x| + C\) and \(\int \frac{1}{x-1} \, dx = \ln|x-1| + C\).
  • Combine these results, accounting for constants of integration, to achieve the final integrated form.
For the given problem, the integration process results in the expression \(x^2 - \ln|x| + \ln|x-1| + C\), where \(C\) represents the constant of integration.

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Most popular questions from this chapter

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