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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 8.1.71b

71. Different Methods
Let I = ∫ (x²)/(x + 1) dx.
b. Evaluate I by first performing long division on the integrand.

Guida verificata passo dopo passo
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Perform long division on the integrand \( \frac{x^2}{x+1} \). Divide \( x^2 \) by \( x+1 \), which gives a quotient of \( x \) and a remainder of \( -x \). Rewrite the integrand as \( x - \frac{x}{x+1} \).
Split the integral \( I \) into two parts: \( I = \int x \, dx - \int \frac{x}{x+1} \, dx \).
Evaluate the first integral \( \int x \, dx \) using the power rule: \( \int x \, dx = \frac{x^2}{2} + C_1 \), where \( C_1 \) is a constant of integration.
For the second integral \( \int \frac{x}{x+1} \, dx \), perform substitution. Let \( u = x+1 \), so \( du = dx \) and \( x = u-1 \). Rewrite the integral as \( \int \frac{u-1}{u} \, du \), which simplifies to \( \int 1 \, du - \int \frac{1}{u} \, du \).
Evaluate the simplified integrals: \( \int 1 \, du = u \) and \( \int \frac{1}{u} \, du = \ln|u| \). Substitute back \( u = x+1 \) to get \( x+1 - \ln|x+1| + C_2 \), where \( C_2 \) is another constant of integration. Combine all results to express the final solution for \( I \).

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Long Division of Polynomials

Long division of polynomials is a method used to divide a polynomial by another polynomial of equal or lower degree. In the context of integration, this technique simplifies the integrand, allowing for easier evaluation of the integral. By dividing the numerator by the denominator, we can express the integrand as a sum of a polynomial and a proper fraction, which can be integrated separately.
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Integration Techniques

Integration techniques refer to various methods used to evaluate integrals, including substitution, integration by parts, and partial fraction decomposition. After simplifying the integrand through long division, the resulting polynomial can be integrated directly, while any remaining proper fraction may require additional techniques for evaluation. Understanding these methods is crucial for effectively solving integrals.
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Definite vs. Indefinite Integrals

Indefinite integrals represent a family of functions whose derivatives yield the integrand, typically expressed with a constant of integration. In this problem, we are evaluating an indefinite integral, which means we will find the antiderivative of the simplified expression. Recognizing the difference between definite and indefinite integrals is essential for correctly interpreting the results of integration.
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