Problem 21 - Calculus: Graphical, Numerical, Algebraic

Question

\(\int \frac{x^{2}+x-1}{x^{2}-x} d x\)

Step-by-Step Solution

Answer

The evaluated integral is \(x + 2x - ln|x-1|+ C\).

Step 1: Polynomial Division

First perform polynomial division to simplify the integral \(\int \frac{x^{2}+x-1}{x^{2}-x} d x\). After dividing \(x^{2}+x-1\) by \(x^{2}-x\) you get \(1+\frac{2x-1}{x^{2}-x}\), therefore the integral becomes \(\int 1+\frac{2x-1}{x^{2}-x} dx\).

Step 2: Decompose Integral

The integrand function is now in a form where it can be decomposed into more manageable sub-integrals. Decomposing the integral yields \(\int 1 dx + \int \frac{2x-1}{x^{2}-x} dx\).

Step 3: Simplify Integral terms

The second integral term \(\int \frac{2x-1}{x^{2}-x} dx\) can be simplified into \(\int 2 dx - \int \frac{1}{x-1} dx\). Therefore, our whole integral becomes \(\int 1 dx + \int 2 dx - \int \frac{1}{x-1} dx\).

Step 4: Calculate Each Integral

Calculated each integral separately: \(\int 1 dx = x\), \(\int 2 dx = 2x\), \(\int \frac{1}{x-1} dx = ln|x-1|\).

Step 5: Combine Results

Combine the results of the computed integrals \(\int 1 dx + \int 2 dx - \int \frac{1}{x-1} dx\) to get the final result \(x + 2x - ln|x-1|+ C\).

Key Concepts

Polynomial Division

Before diving into complex integral calculus techniques, the process of polynomial division often helps simplify integrals. Imagine dividing numbers, but instead, we work with polynomials. In this context, polynomial division breaks down a fraction, making it easier to work with.

Involved is dividing two polynomial expressions using a similar procedure to long division. This is useful when the degree of the numerator is equal to or greater than the degree of the denominator.

• Write the polynomial in descending order of powers.
• Divide the first term of the numerator by the first term of the denominator and multiply the entire denominator by this quotient.
• Subtract this product from the numerator and repeat the process with the new numerator until the remainder's degree is less than the denominator's degree.

In our exercise, polynomial division simplifies the integral of overseer expressions. It initially transforms the complex fraction into a simpler integrable form with a quotient and a reduced fraction.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break apart complex rational expressions into simpler fractions. This tool is fundamental in integral calculus, especially when dealing with fractions that result from polynomial division.

Once you have done polynomial division, the result might include a fraction that can be further simplified. Decomposition involves expressing this fraction as a sum of smaller fractions, typically with simpler denominators. This approach helps when finding their integrals separately is more straightforward.

To apply partial fraction decomposition, follow these steps:

• Factor the denominator completely if possible.
• Set up an equation where the original fraction equals the sum of potential fractions, using letter constants as placeholders for unknown numerators.
• Multiply through by the common denominator to clear the fractions.
• Solve for the constants by equating coefficients or substituting values.

After decomposition, each simple fraction can be integrated individually, significantly streamlining the integral solving process.

Definite and Indefinite Integrals

Understanding definite and indefinite integrals is crucial in integral calculus. Both are methods of integrating functions, but they serve different purposes and have different outcomes.

An **indefinite integral** gives the family of all antiderivatives of a function. It does not have specific upper or lower limits and always includes a constant of integration, denoted by **C**.

• The indefinite integral of a function, often noted as \(\int f(x) \, dx\), represents the collection of all possible antiderivatives.
• Example: The indefinite integral \(\int 2 \, dx\) equals \(2x + C\), where \(C\) is an arbitrary constant.

A **definite integral**, on the other hand, calculates the exact net area under a curve within a specified interval:

• Expressed usually as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration.
• Results in a numerical value, representing the net area between the curve and the x-axis from \(a\) to \(b\).

In solving the exercise, indefinite integration is used as the limits of integration are not provided. This gives us a general solution that includes any potential constants of integration.