Problem 17 - College Algebra and Calculus: An Applied Approach

Question

Use the Log Rule to find the indefinite integral. $$ \int \frac{2}{3 x+5} d x $$

Step-by-Step Solution

Answer

The indefinite integral is \(\frac{2}{3} ln |3x + 5| + C\).

Step 1: Identify u

In this integral, let's identify the denominator \(3x+5\) as u. We can use substitution method to make it simpler.

Step 2: Calculate du/dx

Differentiating our u, which is \(3x + 5\) with respect to x, we get \(du/dx = 3\).

Step 3: Solve for dx

Solving for dx we have \(dx = du/3\). This will be substituted into the integral equation.

Step 4: Substitution into the integral

We substitute \(u\) and \(dx\) into the integral. We also substitute the constant 2 by 2/3 to get : \(\int \frac{2}{3 u} du\). We can bring out constant multiples from an integral, which gives us \(\frac{2}{3} \int \frac{1}{ u} du\).

Step 5: Apply the Logarithmic Rule of Integration

Applying the rule \(\int \frac{1}{u} du = ln |u|\), the result of our integral expression becomes: \(\frac{2}{3} ln |u|\). Remember to add the constant of integration, C.

Step 6: Substitute for u

Finally, replace u with the original expression \(3x + 5\), so our final answer is: \(\frac{2}{3} ln |3x + 5| + C\)

Key Concepts

Substitution Method

The substitution method is a powerful technique for evaluating integrals, particularly when faced with integrals that include a composition of functions. Essentially, this method simplifies an integral by making a clever substitution that transforms it into a more manageable form. Here's how it works:- Begin by identifying a part of the integral, often a difficult or complex expression, and assign it a new variable, typically labeled as "\( u \)".- Differentiate the expression with respect to \( x \) to obtain \( \frac{du}{dx} \).- Solving for \( dx \), provides a way to rewrite the integral entirely in terms of \( u \). This involves substituting \( dx \) with \( \frac{du}{a} \) where \( a \) is the derivative of the substitution expression.By making these substitutions, the integral can be transformed into a form that is more straightforward to solve, often leading to simpler functions such as polynomial, exponential, or logarithmic functions. Once the integration is complete, always remember to substitute back to the original variable.

Logarithmic Rule

The logarithmic rule of integration is particularly useful for finding the integrals of functions involving the reciprocal of a variable - 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: - \[ \int \frac{1}{u} \, du = \ln |u| + C \]This rule is extremely handy when the substitution method leaves you with an integral featuring a \( \frac{1}{u} \) term. In practice, logarithmic integration emerges frequently when dealing with rational functions where the degree of the polynomial in the denominator is higher than that in the numerator. Remember, the inclusion of the absolute value is crucial as it accounts for all possible values of \( u \), ensuring the accuracy of the integral's domain.

Indefinite Integral

An indefinite integral, sometimes simply referred to as an antiderivative, is the reverse process of differentiation. It seeks to determine a function whose derivative is the given function. Typically denoted as \( \int f(x) \, dx \), the solution to an indefinite integral will include a constant of integration, \( C \), because differentiation of a constant yields zero and thus does not appear in the original function.- The process of solving indefinite integrals encapsulates several techniques such as substitution, integration by parts, and partial fraction decomposition. - Indefinite integrals are critical in calculus as they provide a family of functions and not just a single function solution. Once the integration is complete, the constant \( C \) highlights the fact that there are infinitely many antiderivatives for a given function, each differing by a constant. Recognizing the nature and techniques of indefinite integration is key to successfully mastering calculus topics.