Problem 38 - Calculus Early Transcendentals

Question

Give the appropriate form of the partial fraction decomposition for the following functions. $$\frac{2}{x\left(x^{2}-6 x+9\right)}$$

Step-by-Step Solution

Answer

Answer: The appropriate form of the partial fraction decomposition for the given function is $$\frac{2}{x(x - 3)^2} = \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}$$, with factors x and (x - 3)^2 in the denominator and unknown constants A, B, and C.

Step 1: Factorize the denominator

Factorize the denominator of the given rational function: $$x(x^2 - 6x + 9) = x(x - 3)^2$$ This shows that the given function has two factors in the denominator: a linear factor x and a quadratic factor (x - 3)^2.

Step 2: Write out the partial fractions

Since we have a linear and a quadratic factor in the denominator, the partial fraction decomposition can be written in the following form: $$\frac{2}{x(x - 3)^2} = \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}$$ where A, B, and C are unknown constants that need to be determined.

Step 3: Result of partial fraction decomposition

The appropriate form of the partial fraction decomposition for the given function is: $$\frac{2}{x(x - 3)^2} = \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}$$ with factors x and (x - 3)^2 in the denominator and unknown constants A, B, and C.

Key Concepts

Rational Function Decomposition

Rational function decomposition, often referred to as partial fraction decomposition, is a technique in algebra to break down complex rational expressions into simpler, more manageable pieces. Think of it as the mathematical equivalent of dismantling a machine to understand its parts better. It's especially useful when dealing with rational functions that have polynomial expressions in the numerator and the denominator.

For example, take the rational function \(\frac{2}{x(x^{2}-6x+9)}\). The process starts by fully factoring the denominator to determine the types of factors we have: linear, repeated linear, or quadratic. Recognizing these factors is crucial because the form of the partial fraction decomposition will depend on them. After factoring, the next step is to express the rational function as the sum of simpler fractions with undetermined coefficients, which are then solved using algebraic methods like equating coefficients or substitution.

Linear and Quadratic Factors

Partial fraction decomposition can be particularly tricky when the denominator of the rational function involves both linear and quadratic factors. Linear factors are first-degree polynomials, such as \(x-a\), while quadratic factors are second-degree polynomials, like \(x^2+bx+c\) and can sometimes be repeated. In our example, the denominator factors into \(x(x-3)^2\), which includes a linear factor \(x\) and a repeated quadratic factor \(x-3\).When setting up the partial fractions for a mix of linear and quadratic (or even higher-order) factors, the quadratic factors will typically contribute terms in the form of \(\frac{A}{(x-b)^2}\), \(\frac{B}{(x-b)^3}\) etc., depending on the repetition of the factor. The key is to include enough terms to match the repetition of each factor, and in the case of non-repeated linear factors, we simply have \(\frac{constant}{linear factor}\).

Calculus Techniques

When studying calculus, partial fraction decomposition becomes a powerful tool, primarily when integrating complex rational functions. It's like turning a winding mountain road into a straightforward highway. This strategy hinges on breaking down complicated fractions into simpler fractions that we can integrate using basic calculus techniques.Typically, after decomposing a rational function, you'll encounter integral forms that are easier to tackle, like the natural logarithm for linear terms, or powers of inverse trigonometric functions for quadratic terms. Each simpler fraction can then be integrated term-by-term, ultimately simplifying the process of finding the antiderivative of the original, more complex function.

Integrating Rational Functions

Integrating rational functions can appear daunting at first, but with partial fraction decomposition, it's like breaking down a big boulder into small, carriable pieces. Once the original function is expressed as a sum of simpler fractions through decomposition, each piece can then be integrated independently.For linear terms in the decomposition, the integral will typically result in a logarithmic function. For quadratic factors, you may end up with an arctangent function if you have irreducible quadratics, or repeated logarithms for repeated linear factors. The particular integral \(\int\frac{2}{x(x-3)^2}dx\) would, after partial fraction decomposition, convert into a sum of integrals involving the terms \(\frac{A}{x}\), \(\frac{B}{x-3}\), and \(\frac{C}{(x-3)^2}\), each of which can be integrated using standard calculus techniques to find the antiderivative.