Problem 43

Question

In Problems $41-46$, assume that $a$ is a positive constant. Find the general antiderivative of the given function. $$ f(x)=\frac{1}{a x+3} $$

Step-by-Step Solution

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Answer

The general antiderivative is $ \frac{1}{a} \ln|a x + 3| + C $.

1Step 1: Understand Antiderivative Concept

The antiderivative of a function is another function whose derivative is the original function. Finding the antiderivative is equivalent to performing the indefinite integral.

2Step 2: Review the Function Form

The given function is $ f(x) = \frac{1}{a x + 3} $. This function resembles the standard form $ \frac{1}{u} $ where the antiderivative is $ \ln|u| $. Use this concept to guide the antiderivation.

3Step 3: Apply the Substitution Method

Let $ u = a x + 3 $. Then the derivative $ du = a \, dx $. This substitution simplifies the integration process.

4Step 4: Rewrite the Integral with u

With the substitution, the function becomes $ \frac{1}{u} $ and $ dx = \frac{du}{a} $. Thus, the integral becomes $ \int \frac{1}{a} \frac{1}{u} \, du $.

5Step 5: Integrate the Simplified Function

The antiderivative of $ \frac{1}{u} $ is $ \ln|u| $. Therefore, the integration becomes $ \frac{1}{a} \ln|u| + C $ where $ C $ is the constant of integration.

6Step 6: Substitute Back Original Variables

Replace $ u $ with $ a x + 3 $ to revert to the original variable. The antiderivative is $ \frac{1}{a} \ln|a x + 3| + C $.

Key Concepts

Substitution MethodIndefinite IntegralLogarithmic IntegrationConstant of Integration

Substitution Method

The substitution method is a powerful tool in calculus for simplifying integrals. It involves changing variables to make integration easier. In our exercise, we started with the function $ f(x) = \frac{1}{a x + 3} $. To simplify, we let $ u = ax + 3 $. This choice is strategic because the derivative $ du $ corresponds to the expression we have in the denominator,
which ultimately changes our integral to a more familiar form. Using substitution:

Identify the part of the integral that can be substituted with a new variable, $ u $.
Take the derivative of $ u $, resulting in $ du = a \, dx $.
Rewrite the integral in terms of $ u $, helping it resemble a simpler, standard form.

This conversion makes the integral more straightforward to evaluate. After integration, you substitute back the original variables to get the final antiderivative.

Indefinite Integral

An indefinite integral is the reverse process of differentiation. It results in the antiderivative of a function, plus a constant $ C $. Solving an indefinite integral means determining a general expression for the antiderivative without specific limits. In this exercise, we transform the original function using substitution to simplify it to $ \int \frac{1}{u} \, du $.
This form is recognizable and easier to integrate. The integral of $ \frac{1}{u} $ is a classic case taught early in calculus courses:

Recognize the integral as similar to $ \frac{1}{u} $, which integrates to $ \ln|u| $.
Don’t forget to include the integration factor $ \frac{1}{a} $ due to the substitution.
The final result is then multiplied by this factor, thus giving us $ \frac{1}{a} \ln|u| + C $.

This highlights the processes involved in indefinite integration, emphasizing understanding rather than memorization.

Logarithmic Integration

Logarithmic integration is the process of integrating functions that result in logarithmic expressions. This often occurs when dealing with rational functions where substitution helps simplify the function to $ \frac{1}{u} $. In this problem, the function $ f(x) = \frac{1}{a x + 3} $ is transformed to fit $ \frac{1}{u} $ by using the appropriate substitution.
The integral of $ \frac{1}{u} $ is a natural logarithm: $ \ln|u| $.Key steps:

Recognize when an integral can be reduced to $ \frac{1}{u} $.
Utilize substitution, letting the denominator be $ u $.
Integrate, resulting in $ \ln|u| $, and revert $ u $ back to the original variable setup.

The outcome in this context is $ \frac{1}{a} \ln|ax + 3| + C $. This specifies how logarithmic integration translates to real-world problems, providing a deeper understanding of calculus.

Constant of Integration

The constant of integration, usually denoted as $ C $, is a crucial part of indefinite integrals.
When integrating a function without limits, there are infinitely many possible antiderivatives that differ by a constant. This is because differentiation of a constant is zero, making it impossible to discern that constant just by integration.With indefinite integrals:

Each antiderivative belongs to a family of functions, differing by some constant $ C $.
Adding $ C $ represents all possible vertical shifts of the antiderivative graph.
It becomes crucial in applications like solving differential equations, where initial conditions can determine $ C $.

In our exercise's solution, $ \frac{1}{a} \ln |ax + 3| + C $, $ C $ is essential to acknowledge the full set of solutions. Understanding this concept broadens problem-solving techniques in calculus.

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