Question

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

Short answer

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

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: $$\int \frac{dx}{x^2 + 6x + 18} = \int \frac{dx}{(x+3)^2 + 9}$$

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: $$\begin{aligned} \int \frac{dx}{(x+3)^2 + 9} &= \int \frac{3\sec^2(\theta) d\theta}{9\tan^2(\theta) + 9}\\ &= \frac{1}{3} \int \frac{\sec^2(\theta) d\theta}{\tan^2(\theta) + 1} \end{aligned}$$

Simplifying the expression

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

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: $$\int \frac{dx}{x^2 + 6x + 18} = \frac{1}{3}\arctan\left(\frac{x+3}{3}\right) + C$$