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Chapter 5: Problem 6

Find the indefinite integral. $$ \int \frac{1}{3 x+2} d x $$

Short Answer

Expert verified
Applying the correct formula for integrating the function, the indefinite integral of \(\frac{1}{3x+2}\) is \(\frac{1}{3}\ln|3x + 2| + C\).

Step by step solution

01

Identify the Correct Form and Formula

The first step is to identify the correct integral form and corresponding formula. This exercise presents a simple logarithmic integral - the function to be integrated is in the form \(\frac{a}{bx + c}\). The integral of such function is given by the formula: \(\frac{1}{b} \ln |bx + c| + C\), where \(C\) is the constant of integration.
02

Applying the Correct Formula

Now substitute the constants in the formula. The integral is \(\frac{1}{3}\ln|3x + 2| + C\).
03

Simplify the Result

The formula has been applied without any need for further simplification. At this point, the integral is solved.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Integration
Logarithmic integration is a technique used in calculus when dealing with integrals that have a specific form involving rational functions. Notably, these integrals look like \(\int \frac{a}{bx + c} \, dx\).
Here, the function inside the integral is in a form that aligns perfectly with a natural logarithmic function after integration. The process links directly to the properties of the natural logarithm, which simplifies the solution of these integrals.
To solve this kind of problem, it's useful to remember the formula:
  • The integral of \(\frac{1}{bx + c}\) is: \(\frac{1}{b} \ln |bx + c| + C\)
This pattern is crucial for recognizing solutions quickly and applying them correctly in calculus.
Constant of Integration
When finding indefinite integrals, the concept of the constant of integration is fundamental. It's symbolized as \(C\) in the solutions.
Why is it necessary? Well, indefinite integrals represent a family of functions, not a single function.
This constant accounts for all possible constant values that could differentiate two functions, ensuring the integral covers every potential version of the original function.
  • Without \(C\), the solution would be incomplete.
  • It stems from the fact that the derivative of a constant is zero.
Thus, when integrating, add \(C\) to denote all possible original functions.
Calculus Formulas
Calculus is built upon a variety of formulas that solve different types of problems. These formulas allow us to perform integration and differentiation effectively. They are the backbone of solving complex calculus equations quickly by recognizing patterns.
For example, in this exercise:
  • We identified the form \(\frac{a}{bx + c}\) and applied the specific integration formula.
These formulas stem from fundamental math principles and are used to transform challenging problems into solvable steps.
Learning these formulas by heart can save time and help solve calculus problems efficiently. Understanding when and how to apply them is equally essential for success in calculus.

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Most popular questions from this chapter

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Is the converse of the second part of Theorem \(5.7\) true? That is, if a function is one-to-one (and therefore has an inverse function), then must the function be strictly monotonic? If so, prove it. If not, give a counterexample.

Find the indefinite integral using the formulas of Theorem \(5.20 .\) $$ \int \frac{x}{9-x^{4}} d x $$

Find the derivative of the function. $$ y=\tanh ^{-1}(\sin 2 x) $$

Find the integral. $$ \int \operatorname{sech}^{2}(2 x-1) d x $$

Prove that if \(f\) has an inverse function, then \(\left(f^{-1}\right)^{-1}=f\).
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