91Ó°ÊÓ

Integration by parts is a technique used to solve calculus integrals that involves expressing an integral in terms of other simpler integrals. This method is usually applied when you have a product of functions inside the integral. The rule can be linked back to the product rule for differentiation.

To apply integration by parts, use the formula:\[ \int u \, dv = uv - \int v \, du \]Here's a basic rundown of the steps:

• First, identify parts of the function you want to integrate as \( u \) and \( dv \).
• Differentiate \( u \) to get \( du \), and integrate \( dv \) to get \( v \).
• Substitute into \( uv - \int v \, du \) to simplify the integral.

In the original problem, we used \( u = \ln x \) and \( dv = dx \), leading to \( du = \frac{1}{x} \, dx \) and \( v = x \). Integration by parts then gave us: \( x \ln x - \int x \, \frac{1}{x} \, dx \). This step revealed a simpler integral, resulting in the solution \( x \ln x - x + C \).

Logarithmic functions are part of the fundamental concepts of mathematics and often appear in calculus problems. They have their own rules and properties which are essential for solving integration problems.

Here are a few key points to remember:

• The natural logarithm, denoted as \( \ln x \), is the logarithm to base \( e \), where \( e \) is approximately 2.718.
• The properties of logarithms, such as \( \ln(ab) = \ln a + \ln b \) and \( \ln(a^b) = b \ln a \), are incredibly useful in both simplifying expressions and integrating.

In the given exercise, the original logarithm was \( \log_6 x \), indicating a base of 6. Using the change of base formula allowed us to work with the natural logarithm instead, facilitating integration by converting \( \log_6 x \) into \( \frac{\ln x}{\ln 6} \).

The change of base formula for logarithms is crucial when working with logs of different bases, especially in calculus. It allows you to convert a logarithm of one base into another base that may be simpler to work with (such as the natural logarithm).

The change of base formula states:\[ \log_b a = \frac{\ln a}{\ln b} \]This formula is particularly useful when integrating logarithms with non-natural bases. Here's how to apply it:

• Identify the base of the logarithm you want to change, as well as the argument of the log.
• Use the formula to express the original logarithm in terms of natural logarithms \( \ln \).

For the problem in question, this transformed \( \log_6 x \) to \( \frac{\ln x}{\ln 6} \). This transformation was critical in solving the integral since it allowed us to work with \( \ln x \) instead, simplifying the integration process.