Problem 32 - College Algebra and Calculus: An Applied Approach

Question

Find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.) $$ \int \frac{\ln 2 x}{x^{2}} d x $$

Step-by-Step Solution

Answer

The indefinite integral \(\int \frac{\ln 2x}{x^{2}} dx \) is equal to \( -\frac{\ln 2}{x} -\frac{1}{x} ln (ln|x|) + C\)

Step 1: Simplify the integrand using logarithmic properties

The expression within the integral \( \frac{\ln 2x}{x^{2}} \) can be split using the properties of logarithms as follows: \( \ln 2x = \ln 2 + \ln x \). Now the integral becomes \( \int \frac{\ln 2 + \ln x}{x^{2}} \,dx = \int (\frac{1}{x^{2}}\ln 2 + \frac{1}{x^{2}}\ln x)\, dx\). The integral of a sum is equal to the sum of two separate integrals, thus it can be rewritten as \( \int \frac{\ln 2}{x^{2}}dx + \int \frac{\ln x}{x^{2}}dx\)

Step 2: Use integration rules

The first integral can be computed using the power rule for integration, which states that the integral of \(x^{n}\) dx is \(\frac{1}{n+1}x^{n+1}\) where n is not equal to -1. Apply the rule to obtain \( -\frac{\ln 2}{x}\). The second integral can be solved using a special rule which states that the integral of \( ln x / x \) dx is \( ln (ln(|x|) ) + C \). Hence, the integral becomes \( -\frac{1}{x} ln (ln(|x|) ) \)

Step 3: Combine the results

Adding up the result of both integrals to get the final answer. Therefore, the solution of the integral is \( -\frac{\ln 2}{x} -\frac{1}{x} ln (ln(|x|) ) + C \), where C is the constant of integration

Key Concepts

Integration Rules

Understanding integration begins with knowing the fundamental integration rules that help find the antiderivative or indefinite integral of a function. These rules are essential as they simplify the process by providing standard solutions for different forms of functions.

• The Power Rule: This is one of the most commonly used rules in integration, which states that the integral of \( x^n \) is \( \frac{1}{n+1}x^{n+1} \) provided that \( n eq -1 \). This rule simplifies to directly finding the antiderivative by incrementing the power of \( x \) and dividing by the new power.

Applying this rule means that if you have a simple polynomial or a power of \( x \), you can quickly integrate it by adjusting the exponent as described. Remember, each integration result must have a constant \( C \) added to represent the family of antiderivatives, given by the fact that derivatives of constants are zero, so constants are "introduced" during integration.

Logarithmic Properties

Logarithmic properties are mathematical tricks and transformations involving logarithms that can simplify the expression under an integral, allowing for easier integration. They rely on the fundamental understanding that logarithms turn division into subtraction and multiplication into addition under certain conditions.

• Property: \( \ln(ab) = \ln a + \ln b \) This means the logarithm of a product is equal to the sum of the logarithms of its factors. It is a property frequently used to split complicated logarithmic expressions into simpler ones.

For instance, when you encounter \( \ln 2x \), you can rewrite it as \( \ln 2 + \ln x \) without changing its value but making the integration process easier. By separating \( \ln 2x \) into two simpler terms, each term can be handled individually in the context of integration, turning a complex integral into more manageable parts.

Integration by Parts

Integration by parts is a powerful technique derived from the product rule of differentiation. It simplifies the process of integrating products of functions that appear complex. Although our original exercise did not strictly require it, understanding this method is beneficial for tackling diverse and more challenging integrals.This technique operates under the formula:\[ \int u \, dv = uv - \int v \, du\]Here, \( u \) and \( dv \) are parts of the original integral \( \int u \, dv \). You choose which part of the function will become \( u \) and which \( dv \) to optimize the integration process.When choosing \( u \), aim for a function that becomes simpler when differentiated, and pick \( dv \) such that it offers an easy integration path. This selection is key in easing the complexity of the integral. Integration by parts often requires applying the method multiple times and is typically used when straightforward integration rules aren’t applicable.