Question

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

Short answer

\( \ln|x| - \ln|x+1| + C \)

Step-by-step solution

Set up the Partial Fractions

To decompose \( \frac{1}{x(x+1)} \) into partial fractions, assume \( \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \).

Clear the Denominator

Multiply through by the common denominator \( x(x+1) \) to eliminate fractions: \( 1 = A(x+1) + Bx \).

Expand and Simplify the Equation

Expand the right-hand side: \( 1 = Ax + A + Bx \). Simplify to get \( 1 = (A+B)x + A \).

Equate Coefficients

Equate the coefficients of like terms: \( A + B = 0 \) and \( A = 1 \).

Solve for Coefficients

From \( A = 1 \), substitute into the first equation: \( 1 + B = 0 \). Therefore, \( B = -1 \).

Write the Decomposition

The decomposition is \( \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1} \).

Integrate Each Term Separately

Integrate both terms: \( \int (\frac{1}{x} - \frac{1}{x+1}) \, dx = \int \frac{1}{x} \, dx - \int \frac{1}{x+1} \, dx \).

Solve the Integrations

These integrals solve to: \( \ln|x| - \ln|x+1| + C \), where \( C \) is the constant of integration.

Key concepts

Integration Techniques

Integration can seem daunting at first glance, but with various techniques, it becomes manageable. One such technique is partial fraction decomposition. It transforms a complex integral into simpler parts. This is especially useful when dealing with rational functions. In our example, \[\int \frac{1}{x(x+1)} d x\], we decompose the expression into simpler fractions. This allows us to integrate each term individually, making the process much simpler and more intuitive. By applying partial fraction decomposition, we reduce challenging calculus problems to simpler forms. This is not only efficient but also broadens our toolbox for diverse integration challenges.
Integrating these simpler fractions often involves basic integrations, such as the natural log function, which are much more approachable. For this example, the result of integrating after decomposition gives us \[ \ln|x| - \ln|x+1| + C \]. This showcases the power of partial fraction decomposition in tackling integration problems.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. These functions appear frequently in calculus and understanding them is vital. In our example, \[\frac{1}{x(x+1)}\], the function is already in a rational form. Rational functions can often be complex due to their form, so breaking them into more digestible parts helps.
Partial fraction decomposition is especially effective here, as it separates a single complicated rational function into simpler, more integrable terms. This disassembly not only aids in easier computation but also enhances comprehension of the underlying function behavior. Simplifying functions like this doesn't just help in integration, but also in analyzing limits, asymptotes, and other calculus concepts.

Coefficient Comparison

A crucial step in partial fraction decomposition is comparing coefficients. This involves ensuring that both sides of a given equation are equal. After decomposing our initial fraction into \[ \frac{A}{x} + \frac{B}{x+1} \], multiplying both sides by the denominator clears the fractions. We then distribute and combine like terms, \[1 = (A+B)x + A\].
At this stage, we compare coefficients on both sides of the equation. From the constant terms, we derive equations like \[A + B = 0\]. Solving these gives us the values for \(A\) and \(B\), crucial for subsequent steps. This method ensures precision in breaking down rational functions and is fundamental to correct decomposition. Coefficient comparison guarantees that both sides of the equation accurately reflect the polynomial behavior of the terms involved.

Calculus Problem-Solving

Calculus problem-solving involves many layers, from understanding theorems to applying techniques like integration. Partial fraction decomposition is a prime example of a tactical approach within this area. By addressing calculus problems step-by-step—from setting up the decomposition, clearing denominators, and using coefficient comparison, to integrating separate terms—we tackle problems systematically. In the given exercise, \( \int \frac{1}{x(x+1)} d x \), we apply these steps methodically.
Problem-solving in calculus is about leveraging these techniques to simplify complexity. It's also about gaining deeper insights into the mathematical structures we work with. Simplifying integrals not only solves the immediate problem but also fosters a deeper understanding of calculus. This makes tackling new and increasingly complex challenges both achievable and stimulating.