Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-36, use integration by parts to evaluate each integral. $$ \int \ln 3 x d x $$

Step-by-Step Solution

Answer

\( x \ln 3x - x + C \)

Step 1: Identify Parts for Integration by Parts

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). For \( \int \ln 3x \, dx \), we choose \( u = \ln 3x \) so that \( du = \frac{1}{3x} \cdot 3 \, dx = \frac{1}{x} \, dx \). We choose \( dv = dx \) so that \( v = x \).

Step 2: Apply the Integration by Parts Formula

Using the integration by parts formula, substitute the identified parts: \[ \int \ln 3x \, dx = x \ln 3x - \int x \left( \frac{1}{x} \right) dx \].

Step 3: Simplify the Integral

Simplify \( \int x \left( \frac{1}{x} \right) dx \) to \( \int 1 \, dx = x \). Therefore, the entire expression becomes: \[ x \ln 3x - \int 1 \, dx = x \ln 3x - x \].

Step 4: Add the Constant of Integration

Finally, add the constant of integration \( C \) to the solution since the exercise involves indefinite integration. The solution to the integral is: \[ x \ln 3x - x + C \].

Key Concepts

Indefinite Integration

Indefinite integration is the process of finding an antiderivative of a function, resulting in another function plus a constant. It's called "indefinite" because the solution includes an arbitrary constant, represented by \( C \). This contrasts with definite integration, where limits are provided, resulting in a specific number.When we perform indefinite integration, we're essentially working backwards from derivatives to recover the original function. Here's why the constant matters:

• Every function has infinitely many antiderivatives, differing only by a constant.
• The constant \( C \) represents this family of solutions.

In our problem, after using integration techniques, we end up with \( x \ln 3x - x \). By adding \( C \), we finalize our expression to \( x \ln 3x - x + C \), acknowledging all possible antiderivatives.

Natural Logarithm

The natural logarithm is denoted by \( \ln \) and commonly appears in integration problems. It is the logarithm to the base \( e \), where \( e \approx 2.71828 \). Unlike regular logarithms, the natural logarithm focuses on growth processes that occur naturally.In integrals, the natural logarithm often appears due to its derivative properties:

• The derivative of \( \ln x \) is \( \frac{1}{x} \).
• This allows us to manipulate integrals using properties linked to exponential growth or decay.

In the exercise, \( \ln 3x \) is handled by decomposing the expression using integration by parts, a method that pairs \( \ln 3x \) with "simpler" functions to make the integration feasible.

Integration Techniques

Integration techniques are strategies that help solve integrals that aren't straightforward. One such technique is integration by parts, which is useful when integrating products of functions.The integration by parts formula is:\[ \int u \, dv = uv - \int v \, du \]This transforms a complex integral into a simpler one by carefully selecting parts of the integrand.Steps to use integration by parts in the given problem:

• Choose functions for \( u \) and \( dv \). Here, \( u = \ln 3x \) and \( dv = dx \).
• Differentiate \( u \) to get \( du \) and integrate \( dv \) to get \( v \).
• Apply the formula to transform and simplify the integral.

Our original integral \( \int \ln 3x \, dx \) becomes easier to solve through strategic selection and application of these parts, demonstrating the power of this technique in handling complex integrals.