Question

Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{7 x^{2}+2 x-3}{(2 x-1)(3 x+2)(x-3)} d x\)

Short answer

Express the integrand as partial fractions, find constants, and integrate each term separately.

Step-by-step solution

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.

Clear the Denominators

Multiply both sides of the equation by the denominator \((2x-1)(3x+2)(x-3)\) to eliminate the fractions: \( 7x^2 + 2x - 3 = A(3x+2)(x-3) + B(2x-1)(x-3) + C(2x-1)(3x+2) \).

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 \( C(2x-1)(3x+2) \): \( C(6x^2 + 4x - 3x - 2) = C(6x^2 + x - 2) \).

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 \( C \).

Perform the Integration

Once \( A, B, \) and \( C \) are determined, integrate each term separately: \( \int \frac{A}{2x-1} \, dx = \frac{A}{2} \ln |2x-1| + C_1 \),\( \int \frac{B}{3x+2} \, dx = \frac{B}{3} \ln |3x+2| + C_2 \),\( \int \frac{C}{x-3} \, dx = C \ln |x-3| + C_3 \).

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 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.

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.

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:

• 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.

By developing these skills, students can tackle a wide array of calculus problems with confidence.