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Chapter 7: Problem 59

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

Short Answer

Expert verified
Based on the given step-by-step solution, create a short-answer question: Q: Evaluate the integral: $$\int \frac{dx}{x^2 + 6x + 18}$$ A: $$\frac{1}{3}\arctan\left(\frac{x+3}{3}\right) + C$$

Step by step solution

01

Completing the square

Let's first complete the square in the denominator. We want to rewrite the quadratic expression \(x^2 + 6x + 18\) as \((x+a)^2 + b\) for some constants a and b. To find a, observe that the coefficient of the linear term is \(6=2\cdot(3)\) and a must be half of it. So \((x+3)^2 = x^2 + 6x + 9,\) and we need to add \(9\) to get \(x^2 + 6x + 18\). Hence, we can rewrite the expression as \((x+3)^2 + 9\). Now, our integral becomes: $$\int \frac{dx}{x^2 + 6x + 18} = \int \frac{dx}{(x+3)^2 + 9}$$
02

Applying trigonometric substitution

Next, we will perform a trigonometric substitution to solve this integral. Set \(x+3 = 3\tan(\theta)\) such that \((x+3)^2 = 9\tan^2(\theta)\). The differential \(dx\) is given by \(dx = 3\sec^2(\theta) d\theta\). Make the substitution: $$\begin{aligned} \int \frac{dx}{(x+3)^2 + 9} &= \int \frac{3\sec^2(\theta) d\theta}{9\tan^2(\theta) + 9}\\ &= \frac{1}{3} \int \frac{\sec^2(\theta) d\theta}{\tan^2(\theta) + 1} \end{aligned}$$
03

Simplifying the expression

Using the identity \(\tan^2(\theta) + 1 = \sec^2(\theta)\), we have: $$\frac{1}{3} \int \frac{\sec^2(\theta) d\theta}{\sec^2(\theta)} = \frac{1}{3} \int d\theta$$
04

Integrating and reverting the substitution

Integrating with respect to \(\theta\) yields: $$\frac{1}{3}\theta + C$$ Now, revert the substitution, \(\theta = \arctan\left(\frac{x+3}{3}\right)\): $$\frac{1}{3}\arctan\left(\frac{x+3}{3}\right) + C$$ Therefore, the integral evaluates to: $$\int \frac{dx}{x^2 + 6x + 18} = \frac{1}{3}\arctan\left(\frac{x+3}{3}\right) + C$$

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Most popular questions from this chapter

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{x^{3}+1}{x\left(x^{2}+x+1\right)^{2}} d x$$

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{2}{x\left(x^{2}+1\right)^{2}} d x$$

Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b],\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 2 and an illustration.

Refer to Theorem 2 and let \(f(x)=\sin e^{x}\) a. Find a Trapezoid Rule approximation to \(\int_{0}^{1} \sin \left(e^{x}\right) d x\) using \(n=40\) subintervals. b. Calculate \(f^{\prime \prime}(x)\) c. Explain why \(\left|f^{\prime \prime}(x)\right|<6\) on \([0,1],\) given that \(e<3\). (Hint: Graph \(\left.f^{\prime \prime} .\right)\) d. Find an upper bound on the absolute error in the estimate found in part (a) using Theorem 2.

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{d x}{1+e^{x}}$$
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