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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.6.87

Evaluate the integrals in Exercises 77–90.
87. ∫(x²+2x-1)/(x²+9) dx

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Start by rewriting the integral: \(\int \frac{x^{2} + 2x - 1}{x^{2} + 9} \, dx\).
Split the integral into separate parts by expressing the numerator in a form related to the denominator: write \(x^{2} + 2x - 1\) as \((x^{2} + 9) + (2x - 10)\), so the integral becomes \(\int \frac{x^{2} + 9}{x^{2} + 9} \, dx + \int \frac{2x - 10}{x^{2} + 9} \, dx\).
Simplify the first integral to \(\int 1 \, dx\), and keep the second integral as \(\int \frac{2x - 10}{x^{2} + 9} \, dx\).
For the second integral, split it into two integrals: \(\int \frac{2x}{x^{2} + 9} \, dx - \int \frac{10}{x^{2} + 9} \, dx\). Use substitution for the first part and recognize the standard form for the second part.
Use substitution \(u = x^{2} + 9\) for \(\int \frac{2x}{x^{2} + 9} \, dx\), and recall that \(\int \frac{1}{x^{2} + a^{2}} \, dx = \frac{1}{a} \arctan \left( \frac{x}{a} \right) + C\) for the second integral.

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