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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 8.1.8

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.
∫ (2 ln(z³)) / (16z) dz

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1
Start by simplifying the integrand. Use the logarithm power rule: \(\ln(z^3) = 3 \ln(z)\), so rewrite the integral as \(\int \frac{2 \cdot 3 \ln(z)}{16z} \, dz\).
Simplify the constants in the integrand: \(\frac{2 \cdot 3}{16} = \frac{6}{16} = \frac{3}{8}\). The integral becomes \(\int \frac{3}{8} \cdot \frac{\ln(z)}{z} \, dz\).
Factor out the constant \(\frac{3}{8}\) from the integral: \(\frac{3}{8} \int \frac{\ln(z)}{z} \, dz\).
Recognize that the integral \(\int \frac{\ln(z)}{z} \, dz\) can be solved using substitution. Let \(u = \ln(z)\), then \(du = \frac{1}{z} dz\), which matches the integrand's differential part.
Rewrite the integral in terms of \(u\): \(\frac{3}{8} \int u \, du\). Then integrate \(u\) with respect to \(u\) using the power rule for integration.

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Properties of Logarithms

Understanding logarithm properties, such as ln(a^b) = b ln(a), allows simplification of expressions involving logarithms. This is essential for rewriting ln(z³) as 3 ln(z), making the integral easier to handle.
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Integration by Substitution

Integration by substitution involves changing variables to simplify an integral. Recognizing parts of the integrand as derivatives of a function helps to choose an appropriate substitution, streamlining the integration process.
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Substitution With an Extra Variable

Basic Integration Rules

Familiarity with basic integration formulas, such as ∫ ln(x)/x dx or power rule integrals, is crucial. These rules help evaluate integrals after simplification or substitution, enabling the calculation of antiderivatives efficiently.
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Basic Rules for Definite Integrals