Problem 38

Question

In Problems $37-42, a, b$, and $c$ are constants and $g(x)$ is a continuous function whose derivative $g^{\prime}(x)$ is also continuous. Use substitution to evaluate each indefinite integral. $$ \int \frac{1}{a x+b} d x $$

Step-by-Step Solution

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Answer

The integral evaluates to $ \frac{1}{a} \ln|ax+b| + C $.

1Step 1: Recognize the Integral Type

The integral given is $ \int \frac{1}{ax+b} \, dx $, which resembles the standard integral form $ \int \frac{1}{u} \, du = \ln|u| + C $. This suggests that substitution will be useful.

2Step 2: Set Up the Substitution

To simplify the integration process, set $ u = ax + b $. Then, calculate $ du $ by differentiating: $ du = a \, dx $.

3Step 3: Adjust for Substitution

Solve for $ dx $ from $ du = a \, dx $, resulting in $ dx = \frac{du}{a} $. Substitute $ u $ and $ dx $ into the integral: $ \int \frac{1}{u} \cdot \frac{du}{a} $.

4Step 4: Simplify the Integral

Take the constant $ \frac{1}{a} $ out of the integral: $ \frac{1}{a} \int \frac{1}{u} \, du $. This simplifies to $ \frac{1}{a} \ln|u| + C $.

5Step 5: Substitute Back in Terms of x

Replace $ u $ with $ ax + b $ to go back to the variable $ x $. The result is $ \frac{1}{a} \ln|ax+b| + C $. This is the final result for the integral.

Key Concepts

Substitution MethodContinuous FunctionsIntegration Techniques

Substitution Method

The substitution method is a vital integration technique that simplifies complex integrals into more manageable forms. Think of it as a way of changing variables to turn a challenging problem into one that is more familiar and easier to solve. Here's how it works:

When you identify the part of the integral that closely matches a known form (e.g., known derivative), that part becomes your new variable.
Choose a substitution, typically denoted as $ u $, which simplifies the integral. For the integral $ \int \frac{1}{ax+b} \, dx $, set $ u = ax + b $.
Calculate $ du $ by differentiating $ u $, giving us $ du = a \, dx $. This relationship helps you adjust the differential $ dx $ to terms of $ du $.

Substitution essentially transforms the problem to a new integral in terms of $ u $, making it easier to integrate. After solving, substitute back the original variable to complete the integral.

Continuous Functions

Continuous functions play an essential role in calculus, particularly in integration. But what does it mean for a function to be continuous?

A function $ g(x) $ is continuous over an interval if you can draw it without lifting your pencil from the paper. In mathematical terms, $ g(x) $ has no breaks, jumps, or holes over its domain.
When dealing with integration, continuity ensures the function behaves "nicely," allowing properties like the Fundamental Theorem of Calculus to be applied effectively.
The exercise specifies $ g(x) $ is continuous, meaning we can reliably evaluate the integrated function and approximate areas under the curve for any given interval.

In simplest terms, continuity is a hallmark of predictability, enabling smoother calculations in integral calculus.

Integration Techniques

Integration techniques are analytical strategies used to find integrals, and they come in handy depending on the form of the function you're dealing with.

Among the various methods, substitution is often used when integrals resemble standard derivatives, thus simplifying the process.
Other techniques include integration by parts, particularly useful for products of functions, and partial fraction decomposition for rational functions.
Choosing the right method can drastically simplify calculations and lead to quick, accurate answers.

Understanding different integration techniques is critical. They empower you to handle a wide array of problems, ensuring you're always equipped with the right tool for the task at hand.

Problem 38

Problem 39

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