Evaluate the integral. \( \int \frac{d x}{\sqrt{441+x^{2}}} \)

To evaluate the integral \( \int \frac{d x}{\sqrt{441+x^{2}}} \), we can use a trigonometric substitution. Notice that we can rewrite the expression under the square root as \( \sqrt{21^2 + x^2} \), indicating that we can let \( x = 21 \tan(\theta) \). First, we find the derivative and substitute: - \( dx = 21 \sec^2(\theta) d\theta \) - Then, we substitute into the integral: \[ \int \frac{21 \sec^2(\theta) d\theta}{\sqrt{441 + (21 \tan(\theta))^2}} = \int \frac{21 \sec^2(\theta) d\theta}{\sqrt{441 + 441 \tan^2(\theta)}} = \int \frac{21 \sec^2(\theta) d\theta}{\sqrt{441(1 + \tan^2(\theta))}} = \int \frac{21 \sec^2(\theta) d\theta}{21 \sec(\theta)} \] This simplifies to: \[ \int \sec(\theta) d\theta \] The integral of secant is a standard result: \[ \int \sec(\theta) d\theta = \ln | \sec(\theta) + \tan(\theta) | + C \] Returning to our substitution, we have \( \tan(\theta) = \frac{x}{21} \), and therefore \( \sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1 + \left(\frac{x}{21}\right)^2} = \frac{\sqrt{441 + x^2}}{21} \). Thus: \[ \sec(\theta) + \tan(\theta) = \frac{\sqrt{441 + x^2}}{21} + \frac{x}{21} = \frac{\sqrt{441 + x^2} + x}{21} \] Putting all the pieces together, we have: \[ \int \frac{d x}{\sqrt{441+x^{2}}} = \ln \left| \frac{\sqrt{441 + x^2} + x}{21} \right| + C \] Finally, we can express the answer in a simpler form: \[ \int \frac{d x}{\sqrt{441+x^{2}}} = \ln \left( \sqrt{441 + x^2} + x \right) + C \] This is the evaluated integral.