In Exercises 67–73, use integration by parts to establish the reduction formula.
∫ (ln x)^n dx = x (ln x)^n - n ∫ (ln x)^(n-1) dx

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Start with the integral \( I_n = \int (\ln x)^n \, dx \), where \( n \) is a positive integer.

Apply integration by parts by choosing \( u = (\ln x)^n \) and \( dv = dx \). Then, compute \( du \) and \( v \): - \( du = n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \) - \( v = x \)

Use the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Substitute the expressions for \( u, v, du \) to get: \[ I_n = x (\ln x)^n - \int x \cdot n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \]

Simplify the integral inside: \[ I_n = x (\ln x)^n - n \int (\ln x)^{n-1} \, dx \]

This gives the reduction formula: \[ \int (\ln x)^n \, dx = x (\ln x)^n - n \int (\ln x)^{n-1} \, dx \]

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It transforms the integral of a product of functions into simpler integrals, using the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely is crucial to simplify the integral effectively.

Reduction Formula

A reduction formula expresses an integral involving a power or parameter n in terms of a similar integral with a lower power or parameter (n-1). It helps solve complex integrals recursively by breaking them down into simpler cases.

Logarithmic Functions and Their Derivatives

Understanding the properties and derivatives of logarithmic functions, especially ln(x), is essential. The derivative of ln(x) is 1/x, which plays a key role when applying integration by parts to integrals involving powers of ln(x).

In Exercises 67–73, use integration by parts to establish the reduction formula.∫ (ln x)^n dx = x (ln x)^n - n ∫ (ln x)^(n-1) dx