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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 8.5.57

23-64. Integration Evaluate the following integrals.
57. ∫ (x³ + 5x)/(x² + 3)² dx

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Start by examining the integral \( \int \frac{x^{3} + 5x}{(x^{2} + 3)^{2}} \, dx \). Notice that the denominator is \( (x^{2} + 3)^{2} \), which suggests a substitution involving \( x^{2} + 3 \) might simplify the integral.
Let \( u = x^{2} + 3 \). Then, compute the differential \( du = 2x \, dx \), which implies \( x \, dx = \frac{du}{2} \). This substitution will help rewrite parts of the integral in terms of \( u \) and \( du \).
Rewrite the numerator \( x^{3} + 5x \) as \( x(x^{2} + 5) \). Using the substitution, express \( x^{3} + 5x = x(x^{2} + 5) = x(u - 3 + 5) = x(u + 2) \). This allows you to write the integral as \( \int \frac{x(u + 2)}{u^{2}} \, dx \).
Replace \( x \, dx \) with \( \frac{du}{2} \) from the substitution step, so the integral becomes \( \int \frac{u + 2}{u^{2}} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{u + 2}{u^{2}} \, du \).
Split the integral into simpler terms: \( \frac{1}{2} \int \left( \frac{u}{u^{2}} + \frac{2}{u^{2}} \right) du = \frac{1}{2} \int \left( \frac{1}{u} + 2u^{-2} \right) du \). Now, integrate each term separately using the power rule and logarithmic integration.

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