Problem 47 - Calculus for Biology and Medicine

Question

We will discuss alternatives to comparing powers of \(x\) for finding the coefficients in a partial fraction expansion when the denominator polynomial has a repeated root. (a) Consider the rational function $$f(x)=\frac{x+3}{(x-1)(x+1)^{2}}$$ whose partial fraction expression is of the form $$f(x)=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}$$ for some set of constants \(A, B, C\) that need to be determined. To calculate these constants put all of the terms over a common denominator. \(\frac{x+3}{(x+1)^{2}(x-1)}=\frac{A(x+1)^{2}+B(x-1)(x+1)+C(x-1)}{(x-1)(x+1)^{2}}\) Explain why $$x+3=A(x+1)^{2}+B(x-1)(x+1)+C(x-1)$$ (b) One method to calculate \(A, B, C\) from \((7.24)\) is to substitute in specific values of \(x\). We showed in this section that some good choices are values that make one or more terms vanish. Show by substituting in \(x=1\) and \(x=-1\) that \(A=1\) and \(C=-1\). (c) Explain why there is no value of \(x\) that will make both the \(A(x+1)^{2}\) and \(C(x-1)\) terms vanish, without causing the \(B(x-1)(x+1)\) term to vanish. (d) Although we cannot isolate the term in \(B\), we can obtain more equations by substituting different values of \(x .\) By letting \(x=0\), show that: $$3=A-B-C$$ and use this equation to calculate \(B\). (e) Use your answers from (a)-(d) to evaluate $$\int \frac{x+3}{(x+1)^{2}(x-1)} d x$$ (f) Use the method given in parts (b) and (d) to find the partial fraction expansion of \(f(x)=\frac{3 x^{2}-2 x-4}{2 x^{2}(1+x)}.\)

Step-by-Step Solution

Answer

Constants are: A = 1, B = -1, C = -1. Integral: \(\ln |x-1| - \ln |x+1| + \frac{1}{x+1} + C\).

Step 1: Understand the Problem

We're given the rational function \(f(x) = \frac{x+3}{(x-1)(x+1)^2}\) and need to express it in the form \(\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2}\). The goal is to find the constants \(A\), \(B\), and \(C\). To do this, equate the numerators after expressing each term with a common denominator. This is represented by the equation \(x+3 = A(x+1)^2 + B(x-1)(x+1) + C(x-1)\).

Step 2: Substitute Specific Values for x

To find \(A\) and \(C\), substitute values of \(x\) that simplify the equation. Substitute \(x=1\) into \(x+3 = A(x+1)^2 + B(x-1)(x+1) + C(x-1)\) to make \(B\) and \(C\) terms vanish. This simplifies to \(4 = 4A\), leading to \(A = 1\). Then substitute \(x = -1\) to make \(A\) and part of \(B\) vanish. This simplifies to \(2 = -2C\), resulting in \(C = -1\).

Step 3: Analyze Remaining Equation

There is no value of \(x\) that makes the \(A(x+1)^2\) and \(C(x-1)\) terms vanish simultaneously without making the \(B(x-1)(x+1)\) term also vanish. Hence we need other methods to find \(B\).

Step 4: Use Additional Values for x

Substitute \(x = 0\) into the equation: \(x+3 = A(x+1)^2 + B(x-1)(x+1) + C(x-1)\). This results in \(3 = A - B - C\). With \(A = 1\) and \(C = -1\), solve for \(B\): \(3 = 1 - B + 1\), so \(B = -1\).

Step 5: Integrate the Expression

Use the constants found: \(A=1\), \(B=-1\), \(C=-1\). The integral is \(\int \left( \frac{1}{x-1} - \frac{1}{x+1} - \frac{1}{(x+1)^2} \right) dx\), which evaluates to \(\ln |x-1| - \ln |x+1| + \frac{1}{x+1} + C\).

Step 6: Apply Method to a New Function

For \(f(x) = \frac{3x^2 - 2x - 4}{2x^2(1+x)}\), use the method to find the partial fraction form: \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{1+x}\). Substitute suitable \(x\) values (e.g., making terms vanish) to solve for constants \(A\), \(B\), and \(C\), following similar steps above.

Key Concepts

Rational Functions

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. These can be quite versatile and are encountered frequently in calculus and algebra. In simpler terms, it's like a complex version of a basic fraction, but with polynomials instead of just numbers. For example, in the exercise, we are dealing with the rational function \(f(x) = \frac{x+3}{(x-1)(x+1)^2}\). Here, the numerator is a polynomial \(x+3\) and the denominator is a product of the polynomial terms \((x-1)(x+1)^2\).

When you want to perform operations such as differentiation or integration on rational functions, it can often be useful to decompose them into simpler parts. This is where partial fraction decomposition comes in handy! It allows you to break down a complicated rational function into a sum of simpler terms, which are easier to work with individually.

Understanding these functions is crucial because they help simplify complex expressions and make solving equations easier. Many times, the key to tackling more advanced math problems is hidden right in understanding how these rational functions behave.

Repeated Roots

When working with rational functions, you may encounter situations where the denominator has repeated roots. This means that a particular factor appears more than once. For instance, in the rational function \(\frac{x+3}{(x-1)(x+1)^2}\), \((x+1)^2\) has a repeated root because the factor \(x+1\) appears twice.

In partial fraction decomposition, repeated roots require special attention. Instead of having just one fraction for \((x+1)\), you will have several terms. Each term will have an increasing power of the repeated root in the denominator. Hence, the expression for the repeated root \((x+1)^2\) will be broken down into two terms: \(\frac{B}{x+1}\) and \(\frac{C}{(x+1)^2}\).

This approach allows us to solve for each coefficient separately. Repeated roots might seem a bit more complicated initially, but understanding them is crucial because they follow a clear pattern that once mastered, can be applied to any similar problem.

Integration of Rational Functions

Integrating rational functions often involves the technique of partial fraction decomposition, especially when dealing with more complicated expressions. By expressing a complex rational function as a sum of simpler fractions, it becomes easier to find the integral.

For example, let's consider the integral \(\int \frac{x+3}{(x-1)(x+1)^2} \, dx\). After applying partial fraction decomposition, assume the form \(\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2}\). Each of these terms can be integrated separately.

• \(\int \frac{1}{x-1}\, dx\) integrates to \(\ln|x-1|\)
• \(\int \frac{1}{x+1}\, dx\) integrates to \(\ln|x+1|\)
• \(\int \frac{1}{(x+1)^2} \, dx\) integrates to \(-\frac{1}{x+1}\)

This decomposition simplifies our work. We now have a set of easier integrals to handle instead of a more complex initial expression.

By mastering this method, the integration of rational functions becomes more straightforward, making it a valuable tool in both calculus and applied mathematics.