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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.5.23

23-64. Integration Evaluate the following integrals.
23. ∫ [3 / ((x - 1)(x + 2))] dx

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Start by expressing the integrand \( \frac{3}{(x - 1)(x + 2)} \) as a sum of partial fractions. Assume it can be written as \( \frac{A}{x - 1} + \frac{B}{x + 2} \), where \(A\) and \(B\) are constants to be determined.
Multiply both sides of the equation by the denominator \( (x - 1)(x + 2) \) to clear the fractions, resulting in \( 3 = A(x + 2) + B(x - 1) \).
Expand the right-hand side to get \( 3 = A x + 2A + B x - B \), then group like terms: \( 3 = (A + B) x + (2A - B) \).
Set up a system of equations by equating the coefficients of corresponding powers of \(x\) on both sides. Since the left side has no \(x\) term, the coefficient of \(x\) must be zero, and the constant term must be 3. So, \( A + B = 0 \) and \( 2A - B = 3 \).
Solve the system for \(A\) and \(B\), then rewrite the integral as \( \int \left( \frac{A}{x - 1} + \frac{B}{x + 2} \right) dx \). Finally, integrate each term separately using the formula \( \int \frac{1}{x - c} dx = \ln|x - c| + C \).

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