Problem 27 - Calculus Single Variable

Question

In each of Exercises \(27-38,\) use the method of partial fractions to calculate the given integral. $$ \int \frac{3 x+1}{x^{2}-1} d x $$

Step-by-Step Solution

Answer

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

Step 1: Factor the Denominator

The first step in partial fraction decomposition is to factor the denominator as much as possible. Here, the denominator is \(x^2 - 1\), which can be factored into \((x-1)(x+1)\).

Step 2: Set Up the Partial Fraction Decomposition

We express the integrand as a sum of simpler fractions: \(\frac{3x+1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1}\) where \(A\) and \(B\) are constants we need to determine.

Step 3: Clear the Denominator

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

Step 4: Expand and Collect Like Terms

Expand the right side: \(A(x+1) + B(x-1) = Ax + A + Bx - B\). Combine like terms: \((A + B)x + (A - B)\).

Step 5: Set Up the System of Equations

By equating the coefficients from both sides, we have:1. \(A + B = 3\)2. \(A - B = 1\)

Step 6: Solve the System of Equations

Add equations 1 and 2 to solve for \(A\):\(2A = 4\) \(A = 2\).Substitute \(A = 2\) into equation 1 to solve for \(B\):\(2 + B = 3\)\(B = 1\).

Step 7: Write the Partial Fraction Decomposition

Substituting \(A = 2\) and \(B = 1\) back, we have: \(\frac{3x+1}{x^2-1} = \frac{2}{x-1} + \frac{1}{x+1}\).

Step 8: Integrate Each Term Separately

Now integrate each term separately: \(\int \frac{2}{x-1} dx + \int \frac{1}{x+1} dx\).The integrals are: \(2\ln|x-1| + \ln|x+1|\).

Step 9: Combine and Add Constant of Integration

Combine the logarithmic terms: \(2\ln|x-1| + \ln|x+1| + C\), where \(C\) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus that deals with the accumulation of quantities and the areas under and between curves. When we talk about integrating a function, we are essentially looking to find the antiderivative or the area under the graph of a function.In simpler terms, if you think of differential calculus as slicing things up, integral calculus is about putting them back together.For example, when faced with integrating \( \int \frac{3x+1}{x^2-1} dx \), the goal is to decompose the fraction into simpler parts that we can integrate easily.
Integrals can be challenging when dealing with complex fractions, but using techniques like algebraic manipulation makes the process smoother.One key method is partial fraction decomposition, which allows for breaking complex fractions into simpler parts that you can manage better and integrate individually.

Factoring Quadratic Expressions

Factoring quadratic expressions is an algebraic technique that involves expressing a quadratic polynomial as a product of simpler polynomials. For example, consider \(x^2 - 1\). This expression can be factored because it follows the pattern of a difference of squares, which is written as \((a-b)(a+b)\). Thus, \(x^2 - 1\) can be expressed as \((x-1)(x+1)\).
Factoring is beneficial in many fields of mathematics, particularly for methods like partial fraction decomposition, where simplifying denominators is crucial.More broadly, by breaking down expressions, factoring allows us to see clearly what happens at certain points of the graph of a function, like intersections or zeroes.With partial fraction decomposition, factoring forms the first critical step that allows for systematic simplification and integration.

Logarithmic Integration

Logarithmic integration refers to the integration of rational functions that results in a logarithmic function as the antiderivative. This often happens when you integrate functions like \( \frac{1}{x} \).In our problem, once the partial fraction decomposition is set up as \( \frac{2}{x-1} + \frac{1}{x+1} \), each term is integrated separately.
The result is the natural logarithm function: \( \int \frac{2}{x-1} dx = 2\ln|x-1| \) and \( \int \frac{1}{x+1} dx = \ln|x+1| \).These mean we're looking for a function that reverses the differentiation process, leading us to the natural logarithm.Logarithmic integration is vital in calculus because logarithmic functions model many real-world phenomena, making this method widely applicable.

System of Equations

A system of equations consists of two or more equations with the same set of variables. In the context of partial fraction decomposition, after expanding and collecting like terms, we engage with systems to solve for unknowns.When we equate coefficients, we derive equations that express relationships between variables.For example, we ended up with the system:

• \(A + B = 3\)
• \(A - B = 1\)

We solve this system to find the values of \(A\) and \(B\), crucial for decomposing the original fraction into integrable terms.
To solve, we can add the equations to eliminate one of the variables or use substitution, leading to solutions \(A=2\) and \(B=1\).Understanding systems of equations is not only useful here, but also in numerous applications across mathematics, where finding the correct values for variables determines further operations or solutions.