Question

Find the indefinite integral (a) using the integration table and (b) using the specified method. Integral \mathrm{Method } $$ \begin{aligned} &\int \frac{1}{x^{2}(x+1)} d x\\\ &\text { Partial fractions } \end{aligned} $$

Short answer

The indefinite integral of \( \frac{1}{x^{2}(x+1)} \) using the method of partial fractions is \( -ln|x| -\frac{1}{x} +ln|x + 1| + C \).

Step-by-step solution

Understand the Integral

The integral we have to find is \( \int \frac{1}{x^{2}(x+1)} dx \). This integral is a form of a rational function, where the degree of the numerator is less than the degree of the denominator.

Apply Partial Fraction Decomposition

We decompose the function in the integral into simpler fractions that can be easily integrated using the method of partial fractions. The denominator \(x^{2}(x+1)\) can be written as \(x.x(x+1)\). Using the method of partial fractions, we rewrite it as \( \frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \). Where A, B and C are constants to be determined.

Find Constants A, B, and C

To solve for A, B, and C, we equate the two expressions and multiply them by the denominator \(x^{2}(x+1)\) to get rid of the fractions. Solving for A, B, and C, we get A = -1, B = 1 and C = 1.

Integrate Each Term Separately

Substituting back the value of A, B and C, the integral becomes \( \int (-\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1}) dx \). Integrating each term separately, using the power rule and the integral of \(1/x\) gives us \( -ln|x| -\frac{1}{x} +ln|x + 1| + C \) where C is the constant of integration.

Key concepts

Integration Table

An integration table is a helpful tool in calculus, used to find the indefinite integrals of functions. It provides a list of functions and their corresponding antiderivatives. Often, these tables are organized in a way that is easy to navigate, allowing you to quickly find the integral form for common functions. Using an integration table can save time when solving integrals, especially for standard forms that appear frequently. When applying an integration table, check to make sure that the function you are working with matches the form given in the table. Keep in mind that some functions may need manipulation before they can fit into a standard form from the table. Sometimes, this involves algebraic manipulation, like factoring or completing the square, to align with the table format. This technique is commonly applied in examination settings where efficiency is key.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational functions into simpler components, making integration easier. This technique is particularly useful when dealing with rational expressions where the degree of the numerator is less than the degree of the denominator. The first step is to express the original rational function \[ \frac{1}{x^2(x+1)} \]as a sum of simpler fractions. You propose a decomposition, like \[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \]where A, B, and C are constants that need to be determined. By equating the original function to this decomposition and then clearing the fractions, you can solve for the constants by substituting convenient values for \( x \) or by equating coefficients. This step often involves straightforward algebraic manipulation.Once these constants are found, each simpler fraction can be integrated individually. The partial fraction decomposition thus reduces the original problem into smaller, manageable integrals that can be tackled using basic integral rules.

Rational Function

A rational function is defined as a quotient of two polynomials. In the exercise, we have the rational function \[ \frac{1}{x^2(x+1)} \]This specific form has the property that the degree of the numerator (which is zero here) is less than the degree of the denominator (which is three here), which often makes partial fraction decomposition a suitable method for integration.The key feature of rational functions that makes them intriguing in calculus is their ability to be decomposed into additive simpler terms, particularly when integrals are involved. In this case, applying partial fraction decomposition simplifies the integration process, by turning a complex rational function into simpler fractions.One practical note, when dealing with rational functions, is to always check if polynomial long division is needed. However, in this case, it wasn't necessary because the numerator's degree was already lower. Understanding how to work with these functions is crucial for solving integrals in calculus and for grasping more advanced mathematical concepts.