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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
Tutti i libri di testoHass 15th EditionCh. 8 - Techniques of IntegrationProblema 8.5.46
Capitolo 8, Problema 8.5.46

Evaluate the integrals in Exercises 39–54.
∫ 1 / ((x¹/³ - 1)√x) dx
(Hint: Let x = u⁶.)

Guida verificata passo dopo passo
1
Start with the integral \( \int \frac{1}{(x^{1/3} - 1) \sqrt{x}} \, dx \). The hint suggests the substitution \( x = u^6 \), so express \( x \) in terms of \( u \).
Calculate \( dx \) in terms of \( du \) by differentiating \( x = u^6 \), which gives \( dx = 6u^5 \, du \).
Rewrite the expressions inside the integral using the substitution: \( x^{1/3} = (u^6)^{1/3} = u^2 \) and \( \sqrt{x} = \sqrt{u^6} = u^3 \). Substitute these into the integral along with \( dx = 6u^5 \, du \).
Simplify the integral by substituting all parts: the denominator becomes \( (u^2 - 1) u^3 \), and the numerator is replaced by \( 6u^5 \, du \). This will simplify the integral to a rational function in terms of \( u \).
After simplification, set up the integral in terms of \( u \) and proceed to integrate using appropriate techniques such as partial fraction decomposition or algebraic manipulation.

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Substitution Method

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. By letting x = u⁶, the integral's complicated expressions involving roots and powers become simpler polynomials in u, making integration straightforward.
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Euler's Method

Handling Radicals and Fractional Exponents

Radicals like √x and fractional exponents such as x^(1/3) can be rewritten as powers with rational exponents. Understanding how to manipulate these expressions is essential for applying substitution and simplifying the integral before integrating.
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Introduction to Exponent Rules

Integration of Rational Functions

After substitution, the integral often reduces to a rational function in terms of u. Knowing how to integrate rational functions, possibly by partial fractions or direct power rule application, is crucial to find the antiderivative.
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Intro to Rational Functions
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