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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 8.4.46

In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.
∫ √(x) / (1 - x³) dx (Hint: Let u = x³/2)

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Start by examining the integral \( \int \frac{\sqrt{x}}{1 - x^{3}} \, dx \). Notice the hint suggests the substitution \( u = x^{3/2} \). Since \( x^{3/2} = (x^{1/2})^{3} \), this substitution will help simplify the expression inside the denominator.
Express \( u = x^{3/2} \) and differentiate both sides with respect to \( x \) to find \( du \) in terms of \( dx \). Recall that \( \frac{d}{dx} x^{3/2} = \frac{3}{2} x^{1/2} \), so \( du = \frac{3}{2} x^{1/2} dx \).
Solve for \( x^{1/2} dx \) from the expression for \( du \): \( x^{1/2} dx = \frac{2}{3} du \). This substitution will allow you to rewrite the integral in terms of \( u \).
Rewrite the integral in terms of \( u \) using the substitutions: \( \sqrt{x} dx = x^{1/2} dx = \frac{2}{3} du \) and \( 1 - x^{3} = 1 - (x^{3/2})^{2} = 1 - u^{2} \). The integral becomes \( \int \frac{\frac{2}{3} du}{1 - u^{2}} \).
Recognize that the integral now has the form \( \int \frac{du}{1 - u^{2}} \), which is suitable for a trigonometric substitution. Use the substitution \( u = \sin \theta \) or \( u = \cos \theta \) to simplify the denominator and proceed with the integration.

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Trigonometric Substitution

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