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Evaluate the integral. $$\int \frac{d x}{16 x^{2}+16 x+5}$$

Short Answer

Expert verified
The integral evaluates to \( \frac{1}{4} \tan^{-1}(4x + 2) + C \).

Step by step solution

01

Completing the square

To evaluate the integral \( \int \frac{d x}{16 x^{2}+16 x+5} \), first complete the square in the denominator. Rewrite \( 16x^2 + 16x + 5 \) as:\[ 16(x^2 + x) + 5 = 16((x + \frac{1}{2})^2 - \frac{1}{4}) + 5 \]Simplifying, this becomes:\[ 16(x + \frac{1}{2})^2 - 4 + 5 = 16(x + \frac{1}{2})^2 + 1 \]
02

Rewrite the integral

With the completed square, rewrite the original integral:\[ \int \frac{d x}{16(x + \frac{1}{2})^2 + 1} \]Factor out the constant 16 from the denominator:\[ \int \frac{d x}{16((x + \frac{1}{2})^2 + \frac{1}{16})} \]This becomes:\[ \frac{1}{16} \int \frac{d x}{((x + \frac{1}{2})^2 + (\frac{1}{4})^2)} \]
03

Recognize the arctan form

The integral \( \frac{1}{16} \int \frac{d x}{((x + \frac{1}{2})^2 + (\frac{1}{4})^2)} \) is in the form of the inverse tangent integral:\[ \int \frac{d u}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C \]Here, \( u = x + \frac{1}{2} \) and \( a = \frac{1}{4} \).
04

Apply the formula

Using the inverse tangent integral formula, we have:\[ \frac{1}{16} \left( \frac{4}{1} \right) \tan^{-1}\left(\frac{x + \frac{1}{2}}{\frac{1}{4}}\right) + C= \frac{1}{4} \tan^{-1}(4(x + \frac{1}{2})) + C \]Simplifying further gives:\[ \frac{1}{4} \tan^{-1}(4x + 2) + C \]
05

Write the final answer

The evaluated integral is:\[ \frac{1}{4} \tan^{-1}(4x + 2) + C \] where \( C \) is the constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
Completing the square is a mathematical technique used to transform quadratic expressions into a perfect square trinomial. This manipulation is often used to simplify integrals, solve quadratic equations, and analyze graph parabolas. In calculus, it's particularly useful for integrals involving quadratic polynomials in the denominator. For the integral \( \int \frac{d x}{16 x^{2}+16 x+5} \), the first step in the solution is to complete the square for \( 16x^2 + 16x + 5 \). You begin by factoring out the common factor from the terms involving \( x \):- Start with \( 16(x^2 + x) + 5 \).- Next, find a number that completes the square: \( (x + \frac{1}{2})^2 \) equals \( x^2 + x + \frac{1}{4} \). - Adjust the equation: \( 16((x + \frac{1}{2})^2 - \frac{1}{4}) + 5 \).By converting the quadratic expression into a perfect square, it's easier to work with in calculus as it reveals an underlying structure that can be manipulated further in the integration process.
Inverse Trigonometric Functions
Inverse trigonometric functions are the inverse operations of the trigonometric functions like sine, cosine, and tangent. They are useful in integration when dealing with expressions that suggest a geometric or rotational interpretation.In the context of the integral \( \int \frac{dx}{((x + \frac{1}{2})^2 + (\frac{1}{4})^2)} \), recognizing that this fits the form \( \int \frac{d x}{x^2 + a^2} \), which is associated with the arctangent function, is crucial:- The formula \( \int \frac{d u}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left( \frac{u}{a} \right) + C \) is used here.- Identify \( u = x + \frac{1}{2} \) and \( a = \frac{1}{4} \). - This structure allows us to integrate the expression elegantly using the arctangent function.The use of inverse trigonometric functions simplifies the process of finding antiderivatives for certain types of rational functions.
Definite and Indefinite Integrals
In calculus, integration is the process of finding the antiderivative or the area under the curve of a function. Definite integrals give a numerical result, representing the accumulated value over an interval, while indefinite integrals yield a family of functions, representing all antiderivatives of a function.In the article's context, we're dealing with an indefinite integral of the form:- The integral \( \int \frac{1}{16} \times \frac{d x}{((x + \frac{1}{2})^2 + (\frac{1}{4})^2)} \) simplifies to an expression with the constant of integration \( C \).- The result of \( \frac{1}{4} \tan^{-1}(4x + 2) + C \) - Here, \( C \) represents the constant of integration in indefinite integrals, - Indefinite integrals express the entire family of possible solutions, highlighting the role of the arbitrary constant \( C \).Understanding definite and indefinite integrals is crucial for calculus students as it forms the foundation for solving many real-world problems, from physics to engineering.

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