91Ó°ÊÓ

Integration by parts is a useful technique for solving integrals, especially when dealing with products of functions. The formula is derived from the product rule of differentiation and is expressed as:

• \( \int u \, dv = uv - \int v \, du \)

To apply this method, a function needs to be split into two parts: \( u \) and \( dv \). These are chosen carefully such that differentiating \( u \) (to get \( du \)) and integrating \( dv \) (to get \( v \)) simplifies the integration process.
For our case \( \int (\ln x)^n \, dx \), a strategic selection is:

• \( u = (\ln x)^n \)
• \( dv = dx \)

Differentiating and integrating each gives \( du = n(\ln x)^{n-1} \cdot \frac{1}{x} \, dx \) and \( v = x \). Plug these into the integration by parts formula to achieve:\[ \int (\ln x)^n \, dx = x (\ln x)^n - n \int (\ln x)^{n-1} \, dx \]This result effectively breaks down the original integral into simpler components.

Integrating natural logarithmic functions, such as \( \int (\ln x)^n \, dx \), can be challenging. However, with a solid understanding of integration by parts, it becomes manageable. The integral of natural logarithm \( \ln x \) itself serves as the foundational case for more complex powers of \( \ln x \).
In our initial steps, for both simplicity and effectiveness, we use integration by parts. Here's how it works:

• Starting with \( \int \ln x \, dx \), let \( u = \ln x \) and \( dv = dx \).
• Then \( du = \frac{1}{x} \, dx \) and \( v = x \), yielding \( \int \ln x \, dx = x \ln x - \int x \frac{1}{x} \, dx \).

Simplifying this results in \( \int \ln x \, dx = x \ln x - x \), which is repeated and expanded for higher powers of natural logarithms through a reduction formula.

The reduction formula is a powerful tool that simplifies complex integral evaluations by breaking them down into basic components. We've derived our reduction formula as:\[ \int (\ln x)^n \, dx = x (\ln x)^n - n \int (\ln x)^{n-1} \, dx \]Let's apply this repeatedly for practical solutions:

• Start with \( n = 3 \) for \( \int (\ln x)^3 \, dx \).
• Apply the reduction:
  • First, \( \int (\ln x)^3 \, dx = x(\ln x)^3 - 3 \int (\ln x)^2 \, dx \)
  • Then simplify \( \int (\ln x)^2 \, dx \) similarly.
• For each step towards \( \int \ln x \, dx \), use previous results to reduce complexity.

This method is combined with organizing terms to systematically solve \( \int (\ln x)^3 \, dx \). Finally, substituting and combining outcomes yields:\[ \int (\ln x)^3 \, dx = x((\ln x)^3 - 3(\ln x)^2 + 6 \ln x - 6) \]This illustrates how reduction formulas transform arduous integrations into elegant solutions.