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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.PE.75
Chapter 7, Problem 7.PE.75

Evaluate the integrals in Exercises 31–78.
75. ∫(from -2 to -1)2dv/(v²+4v+5)

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1
Start by examining the integral \( \int_{-2}^{-1} \frac{2}{v^{2} + 4v + 5} \, dv \). Notice that the denominator is a quadratic expression, so the first step is to complete the square for the quadratic \( v^{2} + 4v + 5 \).
Rewrite the quadratic by completing the square: \( v^{2} + 4v + 5 = (v^{2} + 4v + 4) + 1 = (v + 2)^{2} + 1 \). This simplifies the integral to \( \int_{-2}^{-1} \frac{2}{(v + 2)^{2} + 1} \, dv \).
Make a substitution to simplify the integral further. Let \( u = v + 2 \), which implies \( du = dv \). Change the limits of integration accordingly: when \( v = -2 \), \( u = 0 \); when \( v = -1 \), \( u = 1 \). The integral becomes \( \int_{0}^{1} \frac{2}{u^{2} + 1} \, du \).
Recognize that the integral \( \int \frac{1}{u^{2} + 1} \, du \) corresponds to the inverse tangent function \( \arctan(u) \). Therefore, the integral can be expressed as \( 2 \int_{0}^{1} \frac{1}{u^{2} + 1} \, du = 2 [ \arctan(u) ]_{0}^{1} \).
Evaluate the definite integral by substituting the limits into the inverse tangent function: \( 2 ( \arctan(1) - \arctan(0) ) \). This expression represents the value of the original integral.

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Key Concepts

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

Definite Integrals

A definite integral calculates the net area under a curve between two specific limits. It is represented as ∫ from a to b of f(x) dx, where a and b are the lower and upper bounds. Evaluating definite integrals involves finding the antiderivative and then applying the Fundamental Theorem of Calculus.
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Definition of the Definite Integral

Completing the Square

Completing the square is a technique used to rewrite quadratic expressions in the form (x + p)² + q. This simplifies integration, especially when dealing with quadratic denominators, by transforming the expression into a form that matches standard integral formulas involving arctangent or logarithmic functions.
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Completing the Square to Rewrite the Integrand

Integration of Rational Functions

Integrating rational functions often requires algebraic manipulation such as partial fraction decomposition or substitution. When the denominator is a quadratic that cannot be factored easily, rewriting it by completing the square helps to apply standard integral formulas, like those involving arctan for expressions of the form 1/(x² + a²).
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Intro to Rational Functions
Related Practice
Textbook Question

Evaluate the integrals in Exercises 31–78.

47. ∫(1/r)csc²(1+ln(r))dr

Textbook Question

In Exercises 129–132 solve the initial value problem.

131. x dy - (y + √y)dx = 0, y(1) = 1

Textbook Question

Use l’Hôpital’s Rule to find the limits in Exercises 85–108.

95. lim(x→∞) (√(x² + x + 1) - √(x² - x))

Textbook Question

In Exercises 125–128 solve the differential equation.

125. dy/dx = √y cos(√y)

Textbook Question

Evaluate the integrals in Exercises 31–78.

43. ∫tan(ln v)/v dv

Textbook Question

In Exercises 115 and 116, find the absolute maximum and minimum values of each function on the given interval.

116. y = 10x (2 - ln(x)), (0, e²]"133. Find the absolute maximum value of

f(x) = x^2 * ln(1/x)

and say where it is assumed