Use reduction formulas to evaluate the integrals in Exercises 41–50.
∫ 16x^3 (ln(x))^2 dx

검증된 단계별 안내

Identify the integral to solve: \(\int 16x^{3} (\ln(x))^{2} \, dx\).

Factor out the constant to simplify the integral: \(16 \int x^{3} (\ln(x))^{2} \, dx\).

Use integration by parts, letting \(u = (\ln(x))^{2}\) and \(dv = x^{3} dx\). Then compute \(du\) and \(v\): - \(du = 2 \ln(x) \cdot \frac{1}{x} dx = \frac{2 \ln(x)}{x} dx\) - \(v = \frac{x^{4}}{4}\).

Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\), so write \(16 \left( \frac{x^{4}}{4} (\ln(x))^{2} - \int \frac{x^{4}}{4} \cdot \frac{2 \ln(x)}{x} dx \right)\).

Simplify the integral inside and recognize that the new integral is \(\int x^{3} \ln(x) \, dx\), which can be solved using the reduction formula or repeated integration by parts.

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주요 개념

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

Reduction Formulas

Reduction formulas are recursive relations that express an integral involving a power or function in terms of a simpler integral of the same type. They simplify complex integrals by reducing the exponent or power step-by-step, making evaluation manageable through repeated application.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation, used to integrate products of functions. It transforms the integral of a product into simpler integrals, often essential when dealing with logarithmic functions multiplied by powers of x.

Properties of Logarithmic Functions in Integration

Logarithmic functions, like ln(x), have unique differentiation and integration properties. Understanding how to handle powers of ln(x) during integration, especially when combined with polynomial terms, is crucial for applying reduction formulas and integration by parts effectively.

Use reduction formulas to evaluate the integrals in Exercises 41–50.∫ 16x^3 (ln(x))^2 dx