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Completing the square Evaluate the following integrals. $$\int \frac{d x}{x^{2}+6 x+18}$$

Short Answer

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Question: Evaluate the integral of \(\int \frac{dx}{x^2+6x+18}\). Answer: \(\frac{1}{3}\arctan\left(\frac{x+3}{3}\right) + C\)

Step by step solution

01

Completing the square

First, we need to complete the square for the denominator \(x^2+6x+18\). To do this, we want to create a quadratic expression of the form \((x+a)^2 + b\) such that it is equivalent to the given denominator. We can rewrite the denominator as: \(x^2+6x+18 = (x^2+6x) + 18\) To complete the square: \((x^2 + 6x) = (x + \frac{6}{2})^2 - (\frac{6}{2})^2 = (x + 3)^2 - 3^2 = (x + 3)^2 - 9\) So, the complete square is: \(x^2+6x+18 = (x + 3)^2 - 9 + 18 = (x + 3)^2 + 9\) Now we can write the integral as: $$\int \frac{dx}{x^2+6x+18} = \int \frac{dx}{(x+3)^2+9}$$
02

Recognizing the integral form

This integral is now in the form of an inverse tangent function: $$\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$$ From the rewritten integral, we can identify \(a = 3\): $$\int \frac{dx}{(x+3)^2+9} = \int \frac{dx}{3^2+(x+3)^2}$$
03

Applying the inverse tangent formula

Now we can apply the formula for the inverse tangent function: $$\int \frac{dx}{3^2+(x+3)^2} = \frac{1}{3}\arctan\left(\frac{x+3}{3}\right) + C$$ So, the solution to the given integral is: $$\int \frac{dx}{x^2+6x+18} = \frac{1}{3}\arctan\left(\frac{x+3}{3}\right) + C$$

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Key Concepts

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

Integral Calculus
Integral calculus is a core part of mathematical analysis focusing on the accumulation of quantities and the areas under and between curves. In the context of the given exercise, the task is to find the antiderivative, which is the inverse operation of differentiation. This is a common practice in calculus, particularly in solving problems that deal with areas, volumes, and other concepts that are related to the integral.

The process of evaluating definite integrals involves finding the limit of a sum, while indefinite integrals are concerned with finding a function (the antiderivative) that, when differentiated, yields the original function under the integral sign. The antiderivative is often accompanied by a constant of integration, denoted by 'C', which accounts for any constants that may disappear during differentiation. Understanding integral calculus is essential for applying mathematical concepts in various fields such as physics, engineering, and economics.
Inverse Tangent Function
The inverse tangent function, denoted as \(\arctan(x)\) or \(\tan^{-1}(x)\), is one of the primary inverse trigonometric functions. It provides the angle whose tangent is the given number. The function is critical when one needs to integrate functions that can be related to the form \(\frac{1}{a^2+x^2}\).

In the context of the given exercise, after completing the square, the integral matches the standard form for the inverse tangent integral, allowing us to solve it easily. This inverse trigonometric function is defined for all real numbers, which means it's applicable in a wide range of scenarios, and its output is normally confined to angles in the range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). It's also often used to solve problems involving right triangles and in various scientific applications such as signal processing and engineering design.
Quadratic Expressions
Quadratic expressions are polynomials of the second degree, typically in the form \( ax^2 + bx + c \), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\). Completing the square is a method used to solve or simplify quadratic expressions, and it's particularly useful for integrating functions with quadratics in the denominator, as seen in the exercise.

To complete the square for an expression \( x^2 + bx \), half of the \(b\) term is squared and added and subtracted within the expression to create a perfect square trinomial. This creates an equivalent expression that is easier to manage and can be used in various techniques such as solving quadratic equations, graphing parabolas, or, as in the given problem, integrating a function. Well understanding the structure and manipulation of quadratic expressions is crucial for students as they are widely encountered in algebra, calculus, and other areas of mathematics.

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Most popular questions from this chapter

Exact Trapezoid Rule Prove that the Trapezoid Rule is exact (no error) when approximating the definite integral of a linear function.

Three cars, \(A, B,\) and \(C,\) start from rest and accelerate along a line according to the following velocity functions: $$v_{A}(t)=\frac{88 t}{t+1}, \quad v_{B}(t)=\frac{88 t^{2}}{(t+1)^{2}}, \quad \text { and } \quad v_{C}(t)=\frac{88 t^{2}}{t^{2}+1}$$ a. Which car travels farthest on the interval \(0 \leq t \leq 1 ?\) b. Which car travels farthest on the interval \(0 \leq t \leq 5 ?\) c. Find the position functions for each car assuming each car starts at the origin. d. Which car ultimately gains the lead and remains in front?

For a real number \(a\), suppose \(\lim _{x \rightarrow a^{+}} f(x)=-\infty\) or \(\lim _{x \rightarrow a^{+}} f(x)=\infty .\) In these cases, the integral \(\int_{a}^{\infty} f(x) d x\) is improper for two reasons: \(\infty\) appears in the upper limit and \(f\) is unbounded at \(x=a .\) It can be shown that \(\int_{a}^{\infty} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{\infty} f(x) d x\) for any \(c>a .\) Use this result to evaluate the following improper integrals. $$\int_{1}^{\infty} \frac{d x}{x \sqrt{x-1}}$$

Trapezoid Rule and Simpson's Rule Consider the following integrals and the given values of \(n .\) a. Find the Trapezoid Rule approximations to the integral using \(n\) and \(2 n\) subintervals. b. Find the Simpson's Rule approximation to the integral using \(2 n\) subintervals. It is easiest to obtain Simpson's Rule approximations from the Trapezoid Rule approximations, as in Example \(8 .\) c. Compute the absolute errors in the Trapezoid Rule and Simpson's Rule with \(2 n\) subintervals. $$\int_{1}^{e} \frac{d x}{x} ; n=50$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. More than one integration method can be used to evaluate \(\int \frac{d x}{1-x^{2}}\) b. Using the substitution \(u=\sqrt[3]{x}\) in \(\int \sin \sqrt[3]{x} d x\) leads to \(\int 3 u^{2} \sin u d u\) c. The most efficient way to evaluate \(\int \tan 3 x \sec ^{2} 3 x d x\) is to first rewrite the integrand in terms of \(\sin 3 x\) and \(\cos 3 x\) d. Using the substitution \(u=\tan x\) in \(\int \frac{\tan ^{2} x}{\tan x-1} d x\) leads to \(\int \frac{u^{2}}{u-1} d u\)

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