Problem 71 - Calculus Single Variable

Question

Derive the reduction formula $$ \int \ln ^{n}(x) d x=x \ln ^{n}(x)-n \int \ln ^{n-1}(x) d x $$

Step-by-Step Solution

Answer

The reduction formula is derived using integration by parts and is \( \int \ln^n(x) \, dx = x \ln^n(x) - n \int \ln^{n-1}(x) \, dx \).

Step 1: Choose the method of integration

We will use integration by parts to derive the reduction formula. Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \).

Step 2: Choose functions u(x) and dv

Let \( u = \ln^n(x) \) and \( dv = dx \). This choice allows us to differentiate \( u \) and integrate \( dv \).

Step 3: Differentiate u(x) and integrate dv

Calculate \( du \) and \( v \):- \( du = n \ln^{n-1}(x) \cdot \frac{1}{x} \, dx = \frac{n \ln^{n-1}(x)}{x} \, dx \)- \( v = \int dx = x \)

Step 4: Apply integration by parts formula

Substitute \( u \), \( dv \), \( du \), and \( v \) into the integration by parts formula:\[ \int \ln^n(x) \, dx = x \ln^n(x) - \int x \cdot \frac{n \ln^{n-1}(x)}{x} \, dx \] which simplifies to\[ x \ln^n(x) - n \int \ln^{n-1}(x) \, dx \]

Step 5: Simplify the expression

The simplified expression \( x \ln^n(x) - n \int \ln^{n-1}(x) \, dx \) is the desired reduction formula. This formula expresses \( \int \ln^n(x) \, dx \) in terms of \( \int \ln^{n-1}(x) \, dx \).

Key Concepts

Integration by Parts

Integration by parts is a fundamental technique in calculus integration used to tackle the integration of products of functions. The basic idea is derived from the product rule in differentiation and allows you to transform a complex integral into simpler ones. The formula for integration by parts is given by:\[\int u \, dv = uv - \int v \, du\]Here is how the formula works:

• Choose your functions wisely: You need to decide which part of the integrand will be \( u \) and which part will be \( dv \). Usually, \( u \) is the function that simplifies upon differentiation, while \( dv \) is easily integrable.
• Differentiation and Integration: Calculate \( du \) by differentiating \( u \), and find \( v \) by integrating \( dv \).
• Substitution: Plug \( u \), \( dv \), \( du \), and \( v \) into the integration by parts formula. This will usually transform the integral in question into an easier form to solve.

The smart choice of \( u \) and \( dv \) is crucial for simplifying the problem at hand. In our case, choosing \( u = \ln^n(x) \) ensures that the differentiation follows a predictable pattern, making the overall problem more manageable.

Natural Logarithm

The natural logarithm, often denoted as \( \ln(x) \), is the power to which the constant \( e \) (approximately 2.71828) must be raised to obtain the number \( x \). It is a logarithm that is commonly used in calculus due to its elegant properties:

• Derivatives: The derivative of \( \ln(x) \) is well known to be \( \frac{1}{x} \), making it an ideal candidate for the \( u \) function in integration by parts when dealing with logarithmic expressions.
• Integration: Integrating logarithmic expressions can be complex, but often these calculations can be simplified significantly using known reduction formulas or integration by parts.
• Properties: Logarithms turn multiplication into addition and powers into multiplication, which can simplify differentiation and integration processes dramatically.

In the context of the reduction formula \( \int \ln^n(x) \, dx \), the natural logarithm empowers you to utilize integration by parts effectively, given its simplification when derived.

Calculus Integration

Calculus integration is about finding the antiderivatives of functions. It is a fundamental concept used to calculate areas under curves, among numerous other applications:

• Definite and Indefinite Integrals: Calculus involves both, where indefinite integrals represent a family of functions, and definite integrals compute exact area values.
• Techniques: Integration techniques include basic antiderivatives, substitution, partial fractions, and integration by parts. These methods are tools that help in solving complex integrals that appear in real-world scenarios.
• Reduction Formulas: These are especially useful for iteratively reducing the power of functions, a strategy visible in our task of solving \( \int \ln^n(x) \, dx \). By expressing the integral in terms of \( \ln^{n-1}(x) \), it simplifies solving sequential integrals where powers of logarithms are involved.

Understanding and mastering these integration techniques is crucial, and they become even more potent when you recognize patterns, like the relationship with reduction formulas, which ultimately simplify your process.