Problem 48 - Calculus for Biology and Medicine

Question

$$ \text { In Problems , evaluate each integral. } $$ $$ \int \frac{2 x^{2}+2 x-1}{x^{3}(x-3)} d x $$

Step-by-Step Solution

Answer

\(-\frac{13}{9} \ln |x| - \frac{5}{9x} - \frac{1}{6x^2} + \frac{13}{9} \ln |x-3| + C\)

Step 1: Decompose the Function

To evaluate the integral \( \int \frac{2x^2 + 2x - 1}{x^3 (x-3)} \, dx \), we begin by decomposing the rational function into partial fractions. We express \( \frac{2x^2 + 2x - 1}{x^3 (x-3)} \) as several simpler fractions: \[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-3} \].

Step 2: Setup the Equation

Multiply through by the common denominator \( x^3(x-3) \) to eliminate the denominators, resulting in: \[ 2x^2 + 2x - 1 = A x^2 (x-3) + B x (x-3) + C (x-3) + D x^3 \].

Step 3: Expand and Collect Like Terms

Expand the right side to get: \[ 2x^2 + 2x - 1 = A x^3 - 3A x^2 + B x^2 - 3B x + C x - 3C + D x^3 \]. Then, collect the terms according to powers of \( x \), resulting in: \[ (A + D)x^3 + (-3A + B)x^2 + (-3B + C)x - 3C \].

Step 4: Equate Coefficients for Each Power of x

Match coefficients for each power of \( x \):1. \( A + D = 0 \)2. \( -3A + B = 2 \)3. \( -3B + C = 2 \)4. \( -3C = -1 \)

Step 5: Solve for Constants

Solve the system of equations from Step 4:1. From \( -3C = -1 \), we find \( C = \frac{1}{3} \).2. Substitute \( C = \frac{1}{3} \) into \( -3B + C = 2 \) to get \( B = \frac{5}{9} \).3. Substitute \( B = \frac{5}{9} \) into \( -3A + B = 2 \) to get \( A = -\frac{13}{9} \).4. Substitute \( A = -\frac{13}{9} \) into \( A + D = 0 \) to get \( D = \frac{13}{9} \).

Step 6: Integrate Each Part Separately

Substitute the values of \( A, B, C, \) and \( D \) back into the partial fractions and integrate each term separately:- \( \int \frac{-\frac{13}{9}}{x} \, dx = -\frac{13}{9} \ln |x| + C_1 \)- \( \int \frac{\frac{5}{9}}{x^2} \, dx = -\frac{5}{9x} + C_2 \)- \( \int \frac{\frac{1}{3}}{x^3} \, dx = -\frac{1}{6x^2} + C_3 \)- \( \int \frac{\frac{13}{9}}{x-3} \, dx = \frac{13}{9} \ln |x-3| + C_4 \)

Step 7: Combine Results

The integral evaluates to:\[ -\frac{13}{9} \ln |x| - \frac{5}{9x} - \frac{1}{6x^2} + \frac{13}{9} \ln |x-3| + C \] where \( C \) is the constant of integration.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational functions into simpler fractions that are easier to integrate. This method is particularly useful in calculus when dealing with integrals of rational functions, which are expressions that are ratios of two polynomials.
To perform partial fraction decomposition, you aim to express a given rational function as a sum of simpler fractions. For example, in the integral \( \int \frac{2x^2 + 2x - 1}{x^3 (x-3)} \, dx \), the process involves rewriting the function as a sum of fractions:

• \( \frac{A}{x} \)
• \( \frac{B}{x^2} \)
• \( \frac{C}{x^3} \)
• \( \frac{D}{x-3} \)

Each fraction corresponds to a simpler part of the original complex fraction, making it easier to integrate. Partial fraction decomposition involves setting up an equation with these terms and then determining the constants \( A, B, C, \) and \( D \) by equating coefficients from both sides.

Rational Function Integration

The integration of rational functions, those expressed as the ratio of two polynomials, often utilizes partial fraction decomposition. This method simplifies the integration process by converting the rational function into a sum of simpler terms.
When faced with a rational function like \( \frac{2x^2 + 2x - 1}{x^3 (x-3)} \), the goal is to decompose it into a form that makes integration straightforward. Generally, this process follows several steps:

• First, ensure that the degree of the numerator is less than the degree of the denominator. If it is not, polynomial long division is applied to adjust it.
• Next, decompose the function into partial fractions, as shown with coefficients \( A, B, C, \) and \( D \).
• Finally, integrate each simpler fraction individually. For example, basic fractions like \( \frac{1}{x} \) and \( \frac{1}{x-n} \) can be integrated using logarithmic functions, while powers of \( x \) follow standard power integration rules.

Rational function integration ultimately simplifies the task of finding the antiderivative by using algebraic manipulation before calculus operations.

Integration Techniques

Integration techniques involve various methods to find the antiderivative or integral of functions. For rational functions, common techniques include partial fraction decomposition, substitution, and integration by parts. These methods are often used in tandem to handle more complex integrals.
Key Techniques Include:

• Substitution: This technique involves changing variables to simplify the integral into a more familiar form. It is often used when the integral involves composite functions.
• Integration by Parts: Based on the product rule for differentiation, this technique is useful when integrating the product of two functions. It is particularly handy when one of the functions becomes simpler upon differentiation.
• Partial Fraction Decomposition: As discussed, this breaks down rational functions into simpler parts, streamlining the integration process.

Choosing the best integration technique often depends on the form and complexity of the function at hand. Mastery of these methods broadens one's ability to tackle a wide range of calculus problems efficiently.