Question

For the following exercises, integrate using whatever method you choose.\(\int \frac{3 x^{2}+1}{x^{4}-2 x^{3}-x^{2}+2 x} d x\)

Short answer

The integral is \( \ln|x| - 3\ln|x-1| + 2\ln|x-2| - \ln|x+1| + C \).

Step-by-step solution

Factor the Denominator

First, let's factor the denominator of the integrand, which is \(x^4 - 2x^3 - x^2 + 2x\). Factor by taking \(x\) out as a common factor, giving us \(x(x^3 - 2x^2 - x + 2)\). Next, we need to factor \(x^3 - 2x^2 - x + 2\). Through polynomial division or by using the rational root theorem, it factors as \((x - 1)(x^2 - x - 2)\), and further factoring \(x^2 - x - 2 = (x - 2)(x + 1)\). Thus, the complete factorization of the denominator is \(x(x-1)(x-2)(x+1)\).

Set Up Partial Fraction Decomposition

Express the integrand as a sum of partial fractions: \[ \frac{3x^2 + 1}{x(x-1)(x-2)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-2} + \frac{D}{x+1} \]Here, \(A\), \(B\), \(C\), and \(D\) are constants that we need to determine by multiplying both sides by the common denominator and equating numerators.

Solve for Constants in Partial Fractions

Multiply both sides by the common denominator \(x(x-1)(x-2)(x+1)\) and equate the numerators:\[ 3x^2 + 1 = A(x-1)(x-2)(x+1) + Bx(x-2)(x+1) + Cx(x-1)(x+1) + Dx(x-1)(x-2)\]Substitute suitable values for \(x\) (the roots of the denominator) or compare coefficients to find \(A, B, C,\) and \(D\).This yields: \(A = 1, B = -3, C = 2, D = -1\).

Integrate Each Partial Fraction Component

Integrate each partial fraction separately:\[ \int \frac{1}{x} \, dx = \ln|x| + C_1 \]\[ \int \frac{-3}{x-1} \, dx = -3 \ln|x-1| + C_2 \]\[ \int \frac{2}{x-2} \, dx = 2 \ln|x-2| + C_3 \]\[ \int \frac{-1}{x+1} \, dx = -\ln|x+1| + C_4 \]

Combine the Results

Combine all the integrated results together:\[ \int \frac{3x^2 + 1}{x(x-1)(x-2)(x+1)} \, dx = \ln|x| - 3\ln|x-1| + 2\ln|x-2| - \ln|x+1| + C \]where \(C\) represents the overall integration constant, which includes \(C_1, C_2, C_3, \) and \(C_4\).

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in integration, particularly with rational functions, where the degree of the numerator is less than that of the denominator. This method simplifies complex rational expressions into a sum of simpler fractions. This is particularly helpful when the polynomial in the denominator can be factored into linear or irreducible quadratic factors.

To perform partial fraction decomposition, begin by factoring the denominator of the rational expression, if not already factored. Once the expression is factored, write it as a sum of fractions, each with one of the factors in the denominator:

• Linear factors like \(x - a\) will have a fraction of the form \(\frac{A}{x-a}\).


• Quadratic factors like \(x^2 + bx + c\) will have a fraction of the form \(\frac{Bx + C}{x^2 + bx + c}\).

After setting up the partial fractions, multiply through by the common denominator to clear the fractions and collect like terms on both sides.

Finally, solve for the constants by substituting suitable values of \(x\) or by comparing coefficients, and you'll have a simpler expression ready for integration.

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial expression into a product of its simplest components, or factors. This is crucial before using techniques like partial fraction decomposition. For this, one typically starts by looking for common factors across all terms.

Let's see how it works by considering the polynomial \(x^4 - 2x^3 - x^2 + 2x\). The initial step is to factor out any common factor from all the terms, which is \(x\). This leads to \(x(x^3 - 2x^2 - x + 2)\).

Looking deeper into the resulting polynomial, one can use methods like the Rational Root Theorem to identify potential roots. In the given example, \(x - 1\) turns out to be a factor. By dividing the cubic polynomial by \(x - 1\), the expression further simplifies, ultimately breaking down into linear factors: \(x(x-1)(x-2)(x+1)\).

Identifying these factors transforms the expression into a form that is more approachable for other mathematical processes, such as partial fraction decomposition.

Rational Root Theorem

The Rational Root Theorem is a helpful tool when factoring polynomials, as it aids in predicting the possible rational roots of a polynomial equation. This theorem states that if a polynomial with integer coefficients has a rational root \(\frac{p}{q}\), then \(p\) (the numerator) must be a factor of the constant term, and \(q\) (the denominator) must be a factor of the leading coefficient.

To apply this theorem, list the possible values of \(p\) and \(q\) using the factors of the constant term and the leading coefficient. Then, test these values by substituting them into the polynomial to see if any result in zero. These results will indicate the polynomial's roots.

In our example, using \(x^3 - 2x^2 - x + 2\), we use this theorem to narrow down potential rational roots. Testing these guessed values systematically allows us to eventually determine that \(x - 1\) is indeed a factor. This discovery allows polynomial division to factor the polynomial further, simplifying subsequent calculations.

Logarithmic Integration

Logarithmic integration is a straightforward technique often used in calculus when dealing with integrals involving rational functions of the form \(\frac{A}{x-a}\). The integral of these types of functions typically results in a logarithmic expression.

Consider the expression \(\int \frac{A}{x-a} \, dx\). The solution to this form is drawn from the natural logarithm function, leading to \(A \ln|x-a| + C\), where \(C\) is the constant of integration that arises from indefinite integrals. Each term of the partial fractions can be integrated individually using this technique.

In the context of the exercise we're discussing, after determining the constants for each partial fraction, logarithmic integration is applied. Each partial fraction results in a simple natural log expression, combined to yield the complete solution. This technique is valued for its simplicity and direct application to integrals derived from linear fractions.