Problem 38

Question

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 the indefinite integrals. $$ \int \frac{1}{a x+b} d x $$

Step-by-Step Solution

Verified

Answer

The integral is $ \int \frac{1}{a x+b} \, dx = \frac{1}{a} \ln |ax+b| + C $.

1Step 1: Identify the Substitution

Consider substituting the expression in the denominator. Let $ u = a x + b $. Then, differentiate to find $ du $ in terms of $ dx $. This gives $ du = a \, dx $, or $ dx = \frac{du}{a} $.

2Step 2: Substitution in the Integral

Substitute $ u = ax + b $ and $ dx = \frac{du}{a} $ into the integral. This transforms the integral to $ \int \frac{1}{u} \cdot \frac{1}{a} \, du $. It simplifies to $ \frac{1}{a} \int \frac{1}{u} \, du $.

3Step 3: Solve the Transformed Integral

The integral $ \int \frac{1}{u} \, du $ is a standard integral, which equals $ \ln |u| + C $, where $ C $ is the constant of integration. Therefore, $ \frac{1}{a} \int \frac{1}{u} \, du = \frac{1}{a} (\ln |u| + C) $.

4Step 4: Substitute Back in Terms of x

Replace $ u $ with $ ax + b $ to revert to the original variable. The integral becomes $ \frac{1}{a}(\ln |ax + b| + C) $. This results in $ \frac{1}{a} \ln |ax + b| + \frac{C}{a} $. To simplify, let the constant $ \frac{C}{a} $ be represented as simply $ C $.

5Step 5: Write the Final Answer

The evaluated indefinite integral is $ \frac{1}{a} \ln |ax + b| + C $. This expression provides the antiderivative or the indefinite integral.

Key Concepts

Indefinite IntegralsSubstitution MethodAntiderivative

Indefinite Integrals

In calculus, an indefinite integral is a function that describes the accumulation of a quantity.
It allows us to find a function whose derivative would give the integrand, which is the function inside the integral.
An indefinite integral is expressed as $ \int f(x) \, dx $ and represents the family of functions whose derivative is $ f(x) $.

This family of functions is determined by adding a constant, $ C $, since the derivative of a constant is zero.
The result of an indefinite integral is also called an antiderivative.

Within the context of $ \int \frac{1}{a x+b} \, dx $, our task is to find an indefinite integral, and hence identify a function that when differentiated returns back the original integrand expression.

Substitution Method

The substitution method is a valuable technique to evaluate indefinite integrals when the integrand is a complex expression.
This method often simplifies the integration process by changing the variable of integration.
It involves selecting part of the integrand to replace with a single variable, often denoted as $ u $.

Begin by setting $ u = a x + b $, which identifies a portion of the integrand to substitute.
Differentiating $ u $, we obtain $ du = a \, dx $, which can be rearranged to suggest a substitution strategy: $ dx = \frac{du}{a} $.
Substituting back into the integral simplifies the expression to a more manageable form: $ \int \frac{1}{u} \cdot \frac{1}{a} \, du $.

By implementing substitution, the integral transforms into one that's easier to solve, thereby streamlining the process substantially.

Antiderivative

An antiderivative is a crucial concept in calculus, used primarily to reverse the process of differentiation.
This involves finding a function whose derivative yields the given function.
In the original exercise, solving the integral means finding the antiderivative for $ \frac{1}{a x + b} $.

The standard integral for $ \int \frac{1}{u} \, du $ results in $ \ln |u| + C $.
Re-subsituting $ u $ back into terms of $ x $ gives the antiderivative $ \frac{1}{a} \ln |a x + b| + C $.
Here, $ C $ represents the arbitrary constant of integration, acknowledging the presence of a family of solutions.

Acquiring the antiderivative is key to unlocking various integral problems, providing insight into functions and their behaviors under differentiation.

Problem 38

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Problem 38

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