Problem 14
Question
Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{7 x^{2}+2 x-3}{(2 x-1)(3 x+2)(x-3)} d x\)
Step-by-Step Solution
Verified Answer
Express the integrand as partial fractions, find constants, and integrate each term separately.
1Step 1: Set Up Partial Fractions
The first step is to express the fraction \( \frac{7x^{2}+2x-3}{(2x-1)(3x+2)(x-3)} \) as a sum of simpler fractions called partial fractions. Assume \( \frac{7x^{2}+2x-3}{(2x-1)(3x+2)(x-3)} = \frac{A}{2x-1} + \frac{B}{3x+2} + \frac{C}{x-3} \). Here, \( A, B, \) and \( C \) are constants to be found.
2Step 2: Clear the Denominators
Multiply both sides of the equation by the denominator \((2x-1)(3x+2)(x-3)\) to eliminate the fractions: \( 7x^2 + 2x - 3 = A(3x+2)(x-3) + B(2x-1)(x-3) + C(2x-1)(3x+2) \).
3Step 3: Expand and Simplify
Expand the right-hand side and simplify: - For \( A(3x+2)(x-3) \): \( A(3x^2 - 9x + 2x - 6) = A(3x^2 - 7x - 6) \).- For \( B(2x-1)(x-3) \): \( B(2x^2 - 6x - x + 3) = B(2x^2 - 7x + 3) \).- For \( C(2x-1)(3x+2) \): \( C(6x^2 + 4x - 3x - 2) = C(6x^2 + x - 2) \).
4Step 4: Solve for the Coefficients
Equate the coefficients of like terms on both sides:- For \( x^2 \): \( 3A + 2B + 6C = 7 \)- For \( x \): \( -7A - 7B + C = 2 \)- For the constant term: \( -6A + 3B - 2C = -3 \)Solve these simultaneous equations to find \( A, B, \) and \( C \).
5Step 5: Perform the Integration
Once \( A, B, \) and \( C \) are determined, integrate each term separately: \( \int \frac{A}{2x-1} \, dx = \frac{A}{2} \ln |2x-1| + C_1 \),\( \int \frac{B}{3x+2} \, dx = \frac{B}{3} \ln |3x+2| + C_2 \),\( \int \frac{C}{x-3} \, dx = C \ln |x-3| + C_3 \).
6Step 6: Write the Final Answer
Combine the results from Step 5 to get the integral:\( \int \frac{7 x^{2}+2 x-3}{(2 x-1)(3 x+2)(x-3)} \, dx = \frac{A}{2} \ln |2x-1| + \frac{B}{3} \ln |3x+2| + C \ln |x-3| + K \),where \( K \) is the constant of integration.
Key Concepts
Integration TechniquesRational FunctionsCalculus Problem Solving
Integration Techniques
Integration is a crucial concept in calculus that helps us find the accumulation of quantities. When dealing with complex expressions, like rational functions, special techniques can simplify the integration process.
One such method is **partial fraction decomposition**. This technique breaks down a complicated integrand into simpler fractions, making integration more manageable. By decomposing the rational function into partial fractions, each term becomes easier to integrate individually. This method is particularly useful when the degree of the numerator is less than the degree of the denominator.
The process starts by expressing a given rational function as a sum of fractions with simpler denominators. Once this decomposition is achieved, each fraction can be integrated separately using standard integration rules. This simplification is a powerful tool allowing for the integration of functions that would otherwise be too complex to handle directly.
One such method is **partial fraction decomposition**. This technique breaks down a complicated integrand into simpler fractions, making integration more manageable. By decomposing the rational function into partial fractions, each term becomes easier to integrate individually. This method is particularly useful when the degree of the numerator is less than the degree of the denominator.
The process starts by expressing a given rational function as a sum of fractions with simpler denominators. Once this decomposition is achieved, each fraction can be integrated separately using standard integration rules. This simplification is a powerful tool allowing for the integration of functions that would otherwise be too complex to handle directly.
Rational Functions
Rational functions are quotients of two polynomials. They are central to many areas of calculus and are the base for applying partial fraction decomposition. The function given in the original exercise, \( \frac{7x^2 + 2x - 3}{(2x-1)(3x+2)(x-3)} \), is a **rational function** with a polynomial numerator and a polynomial denominator.
Understanding the structure of rational functions helps in determining how to decompose them into partial fractions. This decomposition relies on the factorization of the denominator. The denominator must be expressed as a product of linear factors (or irreducible quadratic factors, if applicable) to set up corresponding partial fractions.
Each factor contributes to a partial fraction with unknown coefficients, which are solved algebraically. This decomposition breaks the function into simpler bits that are easier to integrate or otherwise work with, thus providing a clearer pathway through calculus problems.
Understanding the structure of rational functions helps in determining how to decompose them into partial fractions. This decomposition relies on the factorization of the denominator. The denominator must be expressed as a product of linear factors (or irreducible quadratic factors, if applicable) to set up corresponding partial fractions.
Each factor contributes to a partial fraction with unknown coefficients, which are solved algebraically. This decomposition breaks the function into simpler bits that are easier to integrate or otherwise work with, thus providing a clearer pathway through calculus problems.
Calculus Problem Solving
Problem solving in calculus often involves breaking down difficult problems into smaller, manageable parts. This is where having a solid understanding of different techniques, like the ones used for **partial fraction decomposition**, becomes invaluable.
One of the primary goals in any calculus problem is to simplify complex expressions. By using decomposition methods, students can transform an intimidating integral into a sum of simpler integrals. This makes the solution process step-by-step and more approachable.
Successful calculus problem solving requires:
One of the primary goals in any calculus problem is to simplify complex expressions. By using decomposition methods, students can transform an intimidating integral into a sum of simpler integrals. This makes the solution process step-by-step and more approachable.
Successful calculus problem solving requires:
- A good grasp of fundamental algebraic operations to manipulate expressions.
- An understanding of integrals and integral calculus techniques.
- The ability to apply appropriate mathematical tools, like partial fractions, to simplify problems.
Other exercises in this chapter
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