Problem 21 - Calculus for Biology and Medicine

Question

In Problems , use partial-fraction decompositon to evaluate each integral. $$ \int \frac{2 x^{2}-3 x+2}{\left(x^{2}+1\right)^{2}} d x $$

Step-by-Step Solution

Answer

The integral is \( 2\tan^{-1}(x) - \frac{3x}{x^2+1} + C \).

Step 1: Identify the Partial Fractions

The integral we need to evaluate is \( \int \frac{2x^2 - 3x + 2}{(x^2 + 1)^2} \, dx \). Notice that the denominator is \((x^2 + 1)^2\). The partial fraction decomposition can be written as \( \frac{2x^2 - 3x + 2}{(x^2 + 1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2} \). Our task is to find \(A\), \(B\), \(C\), and \(D\).

Step 2: Set Up the Equation

Multiply both sides by \((x^2 + 1)^2\) to eliminate the denominators: \[ 2x^2 - 3x + 2 = (Ax + B)(x^2 + 1) + (Cx + D) \]. Expand the equation: \[ (Ax + B)(x^2 + 1) = Ax^3 + Bx^2 + Ax + B \]. So we have \[ Ax^3 + Bx^2 + Ax + B + Cx + D = 2x^2 - 3x + 2 \].

Step 3: Equate Coefficients

Equate the coefficients from both sides of the polynomial: 1. Coefficient of \(x^3\): \(A = 0\)2. Coefficient of \(x^2\): \(B = 2\)3. Coefficient of \(x\): \(A + C = -3\) (Since \(A = 0\), \(C = -3\))4. Constant term: \(B + D = 2\) (Since \(B = 2\), \(2 + D = 2\) therefore \(D = 0\))

Step 4: Substitution of Coefficients

Substitution gives the decomposition: \[ \frac{2x^2 - 3x + 2}{(x^2 + 1)^2} = \frac{2}{x^2 + 1} + \frac{-3x}{(x^2 + 1)^2} \]. We can now integrate each part separately.

Step 5: Integrate Each Part

Integrate \( \frac{2}{x^2 + 1} \) which gives \( 2\tan^{-1}(x) \). For \( \int \frac{-3x}{(x^2 + 1)^2} \, dx \), use substitution \( u = x^2 + 1 \), \( du = 2x \, dx \), giving \( -3 \int \frac{1}{u^2} \, du = \frac{3}{u} + C \) back substituting gives \( \frac{3}{x^2 + 1} \).

Step 6: Complete the Solution

Combine the integrated results to give the final answer: \[ 2\tan^{-1}(x) - \frac{3x}{x^2+1} + C \]. This represents the antiderivative of the original integral.

Key Concepts

Partial-Fraction Decomposition

Partial-fraction decomposition is a method used in calculus to break down rational expressions into simpler fractions that are easier to integrate. This technique is particularly useful when dealing with integrands whose denominator is factored into distinct or repeated linear or quadratic factors. In our exercise, we start by recognizing the structure of the integral division, where the numerator is of a lower degree than the denominator:

• For our integral \( \int \frac{2x^2 - 3x + 2}{(x^2 + 1)^2} \, dx \), the denominator is \((x^2 + 1)^2\), which is a repeated irreducible quadratic factor.
• This indicates the need for a partial-fraction decomposition of the form \( \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2} \).

The method involves identifying coefficients \( A, B, C, \) and \( D \) that satisfy this equation when expanded and equated to the original polynomial numerator:

• After equating coefficients, we solve a system of equations.
• In our case, those yield \( A = 0, B = 2, C = -3, D = 0 \).

This provides decomposed parts that can be integrated individually, simplifying the calculation of the integral.

Polynomial Integration

In calculus, polynomial integration is the process of finding the antiderivative (or integral) of a polynomial function. When using polynomial integration in conjunction with partial-fraction decomposition, you're leveraging one of its key advantages: simplification of complex rational functions into manageable components.

• For a basic polynomial, integrating \( ax^n \) yields \( \frac{a}{n+1}x^{n+1} + C \) where \( n eq -1 \).
• In our example, after applying partial-fraction decomposition to the initial integrand, we get terms such as \( \frac{2}{x^2 + 1} \) and \( \frac{-3x}{(x^2 + 1)^2} \).

The result is two distinct polynomial integrals:

• \( \int \frac{2}{x^2 + 1} \, dx \), which is straightforward and relates to the arctangent function, giving \( 2\tan^{-1}(x) \).
• \( \int \frac{-3x}{(x^2 + 1)^2} \, dx \), using substitution, turns into a simpler integrable form, resulting in \( \frac{3}{x^2 + 1} \).

Through polynomial integration, these elements provide the fundamental antiderivative components that are pieced together to conclude our solution.

Integration Techniques

Mastering integration techniques means knowing how to apply appropriate methods to compute integral problems effectively. Each method suits particular types of functions, and here, two specific methods are demonstrated.
For our integral, we employed the following techniques:

• Partial-Fraction Decomposition: Already discussed, it's the hallmark technique for rational functions.
• Substitution: Ideal for integrals like \( \int \frac{-3x}{(x^2 + 1)^2} \, dx \). The substitution \( u = x^2 + 1 \) simplifies the expression by changing variables. Here, \( du = 2x \, dx \) facilitates direct integration of a rational function in terms of \( u \).

Once we find the antiderivatives from these techniques, the constant of integration \( C \) is added to represent the family of solutions possible from indefinite integrals. Becoming proficient in choosing and applying these techniques helps dissect complicated integrals into more manageable tasks, leading ultimately to the correct and complete solution.