Problem 47 - Calculus Single Variable

Question

Calculate each of the indefinite integrals. $$ \int \frac{3 x^{3}-16 x^{2}+26 x-14}{(x-1)^{2}(x-2)^{2}} d x $$

Step-by-Step Solution

Answer

Decompose using partial fractions, integrate each term, and combine results with integration constant.

Step 1: Determine the Partial Fraction Decomposition Form

To solve the integral \( \int \frac{3x^3 - 16x^2 + 26x - 14}{(x-1)^2(x-2)^2} \, dx \), we start by expressing the rational function as a sum of partial fractions. We decompose it into:\[ \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x-2} + \frac{D}{(x-2)^2} \] where A, B, C, and D are constants to be determined.

Step 2: Set Up the Equation for the Numerator

Multiply both sides by \((x-1)^2(x-2)^2\) to eliminate the denominators:\[ 3x^3 - 16x^2 + 26x - 14 = A(x-1)(x-2)^2 + B(x-2)^2 + C(x-1)^2(x-2) + D(x-1)^2 \]

Step 3: Solve for Constants Using Specific Values

Choose values for \(x\) to simplify the equation and solve for A, B, C, and D systematically.1. Let \(x = 1\), solving gives the value for D.2. Let \(x = 2\), solving gives the value for B.3. Substitute other values or simplify coefficient comparison to find A and C.

Step 4: Integrate Each Partial Fraction

Once coefficients are found, integrate each term separately:\[ \int \frac{A}{x-1} \, dx, \quad \int \frac{B}{(x-1)^2} \, dx, \quad \int \frac{C}{x-2} \, dx, \quad \text{and} \quad \int \frac{D}{(x-2)^2} \, dx \] These integrals result in terms involving natural logarithms and polynomial terms.

Step 5: Combine and Write the General Solution

Combine the integrals to express the solution to the original integral in terms of \( x \), using the values for A, B, C, and D found previously. Don't forget to add the constant of integration \( C \):\[ \int \frac{3x^3 - 16x^2 + 26x - 14}{(x-1)^2(x-2)^2} \, dx = A\ln|x-1| - \frac{B}{x-1} + C\ln|x-2| - \frac{D}{x-2} + C \]

Key Concepts

Indefinite Integrals

Indefinite integrals play a vital role in calculus. They represent the reverse of differentiation. When given a function, calculating its indefinite integral essentially means finding a function whose derivative is the original function. In symbolic terms, the indefinite integral of a function \( f(x) \) is given by:

• \( \int f(x) \, dx = F(x) + C \)

Where \( F(x) \) is the antiderivative of \( f(x) \) and \( C \) is the constant of integration.
Indefinite integrals don't have specific numerical values, unlike definite integrals. Instead, they provide a family of functions with an added constant \( C \). This reflects the fact that adding or subtracting any constant will produce a function with the same derivative.
Understanding indefinite integrals is pivotal for solving problems involving accumulation, like computing areas or solving differential equations.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. They can express complex relationships in a manageable form. These functions are foundational in fields like engineering and physics and often appear in integration problems.
When we want to integrate a rational function, one effective tool is partial fraction decomposition. This technique simplifies the rational expression into a sum of simpler fractions. These decomposed fractions can often be integrated more straightforwardly.
In essence, recognizing and decomposing rational functions allows us to tackle problems that initially seem too complicated for traditional integration techniques.

Integration Techniques

Integration is a cornerstone of calculus, involving multiple techniques to solve different types of problems. One powerful method is partial fraction decomposition. This technique is particularly useful when dealing with rational functions.

• Decompose the rational function into simpler fractions using constants \( A, B, C, \) and more.
• Eliminate denominators by multiplying through by the common denominator.
• Solve for constants by setting \( x \) to specific values or equating coefficients.

Once the fractions are decomposed, integrating becomes manageable. Each fraction can be integrated using basic methods, resulting in solutions that involve logarithms and polynomial expressions. This technique not only simplifies the integration process but also broadens our approach to solving complex integral problems effectively.
Mastering these techniques involves practice and understanding the underlying principles, enabling us to dissect complicated expressions efficiently.