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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 8.5.20

In Exercises 17–20, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (x² dx) / ((x - 1)(x² + 2x + 1))

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First, recognize that the denominator can be factored further. Notice that \(x^{2} + 2x + 1\) is a perfect square trinomial, so rewrite it as \((x + 1)^{2}\). Thus, the integral becomes \(\int \frac{x^{2}}{(x - 1)(x + 1)^{2}} \, dx\).
Set up the partial fraction decomposition for the integrand. Since the denominator has a linear factor \((x - 1)\) and a repeated linear factor \((x + 1)^{2}\), express the fraction as: \(\frac{x^{2}}{(x - 1)(x + 1)^{2}} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^{2}}\).
Multiply both sides of the equation by the denominator \((x - 1)(x + 1)^{2}\) to clear the fractions, resulting in: \(x^{2} = A(x + 1)^{2} + B(x - 1)(x + 1) + C(x - 1)\).
Expand the right-hand side and collect like terms in powers of \(x\). This will give you a polynomial equation in \(x\) where the coefficients of corresponding powers of \(x\) on both sides must be equal. Use this to set up a system of equations to solve for \(A\), \(B\), and \(C\).
Once you find the values of \(A\), \(B\), and \(C\), rewrite the integral as the sum of simpler integrals: \(\int \frac{A}{x - 1} \, dx + \int \frac{B}{x + 1} \, dx + \int \frac{C}{(x + 1)^{2}} \, dx\). Then, integrate each term separately using standard integral formulas.

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