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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.2.41

Evaluate the integrals in Exercises 39–56.
41. ∫2y dy/(y²-25)

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1
Identify the integral to solve: \(\int \frac{2y}{y^{2} - 25} \, dy\).
Recognize that the denominator is a quadratic expression that can be factored as \(y^{2} - 25 = (y - 5)(y + 5)\), but before factoring, check if substitution is more straightforward.
Use substitution by letting \(u = y^{2} - 25\), then compute \(du = 2y \, dy\), which matches the numerator times \(dy\) exactly.
Rewrite the integral in terms of \(u\): \(\int \frac{2y}{y^{2} - 25} \, dy = \int \frac{1}{u} \, du\).
Integrate \(\int \frac{1}{u} \, du\) to get \(\ln|u| + C\), then substitute back \(u = y^{2} - 25\) to express the answer in terms of \(y\).

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