50-53. Reduction Formulas Use integration by parts to derive the following reduction formulas: 53. ∫ lnⁿ(x) dx = x lnⁿ(x) - n ∫ lnⁿ⁻¹(x) dx
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Identify the integral to solve: \(\int \ln^{n}(x) \, dx\), where \(n\) is a positive integer power of the natural logarithm function.
Choose parts for integration by parts. Let \(u = \ln^{n}(x)\) and \(dv = dx\). Then, compute \(du\) and \(v\):
- Since \(u = \ln^{n}(x)\), use the chain rule to find \(du = n \ln^{n-1}(x) \cdot \frac{1}{x} \, dx\).
- Since \(dv = dx\), integrate to get \(v = x\).
Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Substitute the expressions for \(u\), \(v\), and \(du\):
\(\int \ln^{n}(x) \, dx = x \ln^{n}(x) - \int x \cdot n \ln^{n-1}(x) \cdot \frac{1}{x} \, dx\).
Simplify the integral inside the subtraction:
\(\int x \cdot n \ln^{n-1}(x) \cdot \frac{1}{x} \, dx = n \int \ln^{n-1}(x) \, dx\).
Rewrite the expression to obtain the reduction formula:
\(\int \ln^{n}(x) \, dx = x \ln^{n}(x) - n \int \ln^{n-1}(x) \, dx\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Integration by Parts
Integration by parts is a technique based on the product rule for differentiation. It transforms the integral of a product of functions into simpler integrals, using the formula ∫u dv = uv - ∫v du. This method is essential for deriving reduction formulas involving powers of logarithmic functions.
Reduction formulas express an integral involving a power or parameter in terms of a similar integral with a lower power or simpler parameter. They simplify complex integrals step-by-step, making it easier to evaluate integrals like ∫lnⁿ(x) dx by relating them to ∫lnⁿ⁻¹(x) dx.
Understanding the natural logarithm function ln(x) and its derivatives is crucial. Since d/dx[ln(x)] = 1/x, this property helps in choosing appropriate parts for integration by parts and manipulating integrals involving powers of ln(x).