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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 8.1.54a

Using different substitutions
Show that the integral
∫((x² - 1)(x + 1))^(-2/3) dx
can be evaluated with any of the following substitutions.
a. u = 1/(x + 1)
What is the value of the integral?

Guida verificata passo dopo passo
1
Start with the integral \( \int ((x^2 - 1)(x + 1))^{-2/3} \, dx \). First, simplify the expression inside the integral. Notice that \( x^2 - 1 = (x - 1)(x + 1) \), so the integrand becomes \( ((x - 1)(x + 1)^2)^{-2/3} \).
Rewrite the integrand as \( (x - 1)^{-2/3} (x + 1)^{-4/3} \) by distributing the exponent \( -2/3 \) to each factor inside the parentheses.
Use the substitution \( u = \frac{1}{x + 1} \). Then, express \( x + 1 \) and \( dx \) in terms of \( u \): \( x + 1 = \frac{1}{u} \) and \( dx = -\frac{1}{u^2} du \) (found by differentiating \( u = \frac{1}{x + 1} \) with respect to \( x \)).
Rewrite \( x - 1 \) in terms of \( u \) using \( x = \frac{1}{u} - 1 \), so \( x - 1 = \frac{1}{u} - 2 \). Substitute these expressions into the integrand and the differential \( dx \) to rewrite the entire integral in terms of \( u \).
Simplify the resulting integral in \( u \) and then integrate using appropriate techniques (such as power rule or further substitution). After integration, substitute back \( u = \frac{1}{x + 1} \) to express the answer in terms of \( x \).

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Substitution Method in Integration

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. By letting u equal a function of x, we rewrite the integral in terms of u and du, making it easier to evaluate. This technique is especially useful when the integral contains composite functions.
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Algebraic Manipulation of Integrands

Before applying substitution, it is important to simplify or rewrite the integrand using algebraic identities or factorization. This helps identify suitable substitutions and reduces the complexity of the integral. For example, expressing (x² - 1)(x + 1) in a simpler form can clarify the substitution process.
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Integration of Power Functions

Integrals involving expressions raised to fractional powers, such as (-2/3), require understanding how to integrate power functions. After substitution, the integral often reduces to a form ∫u^n du, which can be integrated using the power rule: ∫u^n du = u^(n+1)/(n+1) + C, provided n ≠ -1.
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