Problem 16 - College Algebra and Calculus: An Applied Approach

Question

Use the Log Rule to find the indefinite integral. $$ \int \frac{1}{6 x-5} d x $$

Step-by-Step Solution

Answer

The indefinite integral \( \int \frac{1}{6x - 5} dx \) = \( \frac{1}{6} \ln |6x - 5| + C \)

Step 1: Identifying u

Here, we identify that the \( u \) in \( \frac{1}{u} \) is \( (6x-5) \).

Step 2: Define differential of u, du

We propose a du so that we can express dx as equivalent du. The derivative of \( u \) (or \( (6x-5) \)) is \( 6 dx = du \). So, \( dx = \frac{du}{6} \).

Step 3: Substitute and Simplify

Substitute \( (6x-5) \) with \( u \) and \( dx \) with \( \frac{du}{6} \) in the integral. This gives us \( \int \frac{1}{u} \cdot \frac{du}{6} \) which simplifies to \( \frac{1}{6} \int \frac{1}{u} du \).

Step 4: Integration Rule Application

Now, using the rule of integration \( \int \frac{1}{u} du = \ln |u| + C \), here \( u = (6x - 5) \), we can calculate the integral as \( \frac{1}{6} \ln |6x - 5| + C \).

Step 5: Write the Final Solution

The final solution becomes \( \frac{1}{6} \ln |6x - 5| + C \), where \( C \) is the constant of integration.

Key Concepts

Log Rule

The Log Rule is a crucial concept in integral calculus, especially when dealing with integrals of the form \( \int \frac{1}{u} \, du \). This rule states that the integral of \( \frac{1}{u} \) with respect to \( u \) is the natural logarithm of the absolute value of \( u \), plus a constant of integration:

• \( \int \frac{1}{u} \, du = \ln |u| + C \)
• It is important because it simplifies the process of integration for certain types of functions.

In the example given, once \( u = 6x - 5 \) is identified, the log rule allows the indefinite integral to be expressed as \( \frac{1}{6} \ln |6x - 5| + C \). This approach transforms a potentially complex integral into a form that is straightforward to compute.

Substitution Method

The Substitution Method is a powerful tool in calculus for simplifying integrals. It transforms an integral into an easier one by changing variables. The steps involve:

• Identifying a function inside the integral to replace with a single variable \( u \).
• Finding the derivative of that function, which helps to express \( dx \) in terms of \( du \).

In our problem, we used the substitution \( u = 6x - 5 \). This choice made \( du = 6 \, dx \), effectively allowing us to express \( dx \) as \( \frac{du}{6} \). This way, the original integral \( \int \frac{1}{6x-5} \, dx \) is transformed into \( \frac{1}{6} \int \frac{1}{u} \, du \), which is much simpler to solve using the log rule.

Differentiation

Differentiation, or finding the derivative, involves calculating the rate at which a function is changing at any given point. In the context of the substitution method, differentiation is used to transform \( dx \) into a different form that matches our new substitution variable \( u \). With our initial substitution \( u = 6x - 5 \), we differentiate to find \( du \):

• The derivative \( \frac{d}{dx}(6x - 5) = 6 \), so \( du = 6 \, dx \).
• This allows us to express the original \( dx \) in terms of \( du \) - namely, \( dx = \frac{du}{6} \).

This relationship is essential for changing variables correctly within the integral.

Integral Calculus

Integral Calculus is the branch of calculus focused on finding integrals, that is, functions given their derivatives. It is crucial for calculating areas under curves, among other applications. Here, indefinite integration refers to the integration of a function without specific upper and lower bounds for the integral. When performing indefinite integration, the goal is to find the antiderivative (or integral) that describes a family of functions. Each function in the family is shifted vertically by the constant of integration \( C \).

• The notation \( \int f(x) \, dx \) suggests finding a function whose derivative is \( f(x) \).
• The result of indefinite integration includes \( + C \) to represent all possible vertical shifts of the antiderivative.

In the given exercise, integral calculus helps us calculate \( \int \frac{1}{6x-5} \, dx \) by applying the Log Rule and using the substitution \( u = 6x - 5 \), ultimately leading to the solution \( \frac{1}{6} \ln |6x - 5| + C \).