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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.2.70

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

Guida verificata passo dopo passo
1
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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