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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.5.46

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

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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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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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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