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Magazine Chapter 6: Problem 49
Find each integral by whatever means are necessary (either substitution or tables). $$ \int \frac{1}{2 x+6} d x $$
Short Answer
Expert verified
\( \frac{1}{2} \ln |2x + 6| + C \)
Step by step solution
01
Identify the form of the integral
The given integral is \( \int \frac{1}{2x+6} \, dx \). This integral has the standard form of \( \int \frac{1}{ax+b} \, dx \), which can be solved using a natural logarithmic integration method.
02
Determine the substitution
To facilitate the integration process, we can make the substitution \( u = ax + b \). For this problem, let \( u = 2x + 6 \). Then, calculate \( \frac{du}{dx} = 2 \), or \( dx = \frac{1}{2} du \).
03
Substitute into the integral
Replace \( 2x+6 \) with \( u \) and \( dx \) with \( \frac{1}{2} du \) in the integral. The integral becomes \( \int \frac{1}{u} \cdot \frac{1}{2} \, du \), which simplifies to \( \frac{1}{2} \int \frac{1}{u} \, du \).
04
Integrate the new integral
The integral \( \int \frac{1}{u} \, du \) is a standard logarithmic integral, which results in \( \ln |u| + C \), where \( C \) is the constant of integration. Thus, \( \frac{1}{2} \int \frac{1}{u} \, du \) becomes \( \frac{1}{2} \ln |u| + C \).
05
Substitute back for \( u \)
Replace \( u \) with the original expression \( 2x + 6 \) to revert back to the variable \( x \). The solution to the integral is \( \frac{1}{2} \ln |2x + 6| + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique in calculus for simplifying the process of integration. It is especially useful when dealing with functions that can be transformed into a more straightforward form. In essence, the method aims to change the variable of integration to make the integral easier to solve.
Consider the integral \( \int \frac{1}{2x+6} \, dx \). Direct integration seems complex, so the substitution method can significantly simplify it. Here's how it works:
Consider the integral \( \int \frac{1}{2x+6} \, dx \). Direct integration seems complex, so the substitution method can significantly simplify it. Here's how it works:
- Choose a substitution: Here, we select \( u = 2x + 6 \), turning the integral into a form that is easier to handle.
- Calculate the differential: With \( u = 2x + 6 \), we have \( \frac{du}{dx} = 2 \). Rearranging gives \( dx = \frac{1}{2} du \).
- Substitute and rewrite: After substituting into the original integral, it becomes \( \int \frac{1}{u} \cdot \frac{1}{2} \, du \), simplifying to \( \frac{1}{2} \int \frac{1}{u} \, du \).
Logarithmic Integration
Logarithmic integration is a method used when the integrand is in the form \( \frac{1}{u} \). This form is particularly noteworthy because the antiderivative is directly related to the natural logarithm function.
Once we have changed the variable using substitution, our problem transitions into a logarithmic form. For example, after substitution, the integral \( \int \frac{1}{2x+6} \, dx \) turns into \( \frac{1}{2} \int \frac{1}{u} \, du \). This is where logarithmic integration shines:
Once we have changed the variable using substitution, our problem transitions into a logarithmic form. For example, after substitution, the integral \( \int \frac{1}{2x+6} \, dx \) turns into \( \frac{1}{2} \int \frac{1}{u} \, du \). This is where logarithmic integration shines:
- Recognize the form: The integral \( \int \frac{1}{u} \, du \) is a standard form leading to \( \ln |u| + C \), where \( C \) represents the constant of integration.
- Integrate: By recognizing \( \frac{1}{2} \int \frac{1}{u} \, du \) as a logarithmic integral, we find its solution to be \( \frac{1}{2} \ln |u| + C \).
Definite and Indefinite Integrals
In calculus, integrals are categorized into two primary types: definite and indefinite. These serve different purposes and have distinct properties.
- Indefinite Integrals: The integral we solved in the original exercise \( \int \frac{1}{2x+6} \, dx \) is an indefinite integral. An indefinite integral represents a family of functions and includes a constant of integration \( C \). It provides the antiderivative of a function.
- Definite Integrals: Unlike indefinite integrals, definite integrals compute the area under a curve between two specified limits. A definite integral provides a specific numerical value, not just a formula, by evaluating the antiderivative at the upper and lower bounds and subtracting the results.
