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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.1.12
Chapter 8, Problem 8.1.12

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.
∫₋₁³ (4x² - 7) / (2x + 3) dx

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Start by examining the integrand \( \frac{4x^{2} - 7}{2x + 3} \). Since the degree of the numerator (2) is greater than the degree of the denominator (1), perform polynomial long division to simplify the integrand.
Divide \(4x^{2} - 7\) by \(2x + 3\) to express the integrand as a polynomial plus a proper rational function. This will give you an expression of the form \(Q(x) + \frac{R(x)}{2x + 3}\), where \(Q(x)\) is a polynomial and \(R(x)\) is a polynomial of degree less than 1.
Rewrite the integral as the sum of two integrals: \(\int_{-1}^{3} Q(x) \, dx + \int_{-1}^{3} \frac{R(x)}{2x + 3} \, dx\). This breaks the problem into simpler parts.
For the integral involving \(\frac{R(x)}{2x + 3}\), use substitution. Let \(u = 2x + 3\), then compute \(du = 2 \, dx\) or \(dx = \frac{du}{2}\). Change the limits of integration accordingly from \(x\) to \(u\).
Integrate both parts separately: the polynomial integral is straightforward, and the integral involving \(\frac{1}{u}\) will result in a natural logarithm function. After integrating, substitute back to the original variable \(x\) and evaluate the definite integral using the given limits.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Polynomial Division

When integrating a rational function where the numerator's degree is equal to or higher than the denominator's, polynomial division simplifies the integrand into a polynomial plus a proper fraction. This makes the integral easier to evaluate by breaking it into simpler parts.
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Substitution Method

Substitution involves changing variables to simplify an integral, especially when the integrand contains a composite function. By letting u equal a part of the integrand (like the denominator), the integral can be transformed into a more straightforward form.
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Definite Integral Evaluation

After finding the antiderivative, evaluating a definite integral requires applying the Fundamental Theorem of Calculus by substituting the upper and lower limits into the antiderivative and subtracting. This yields the exact area under the curve between the given bounds.
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Definition of the Definite Integral
Related Practice
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Textbook Question

Use the table of integrals at the back of the text to evaluate the integrals in Exercises 1–26.

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