To solve the integral of $\frac{\sqrt{81 + x^2}}{x}$, we can use a substitution method. Let's follow these steps:

Step-by-step solution:

Given: $I = \int \frac{\sqrt{81 + x^2}}{x} \, dx$

Step 1: Substitution

We can use the substitution: $x = 9 \tan(\theta)$ Then, $dx = 9 \sec^2(\theta) \, d\theta$ and using the Pythagorean identity, we have: $x^2 = 81 \tan^2(\theta) \quad \text{so} \quad 81 + x^2 = 81(1 + \tan^2(\theta)) = 81 \sec^2(\theta)$

Thus, $\sqrt{81 + x^2} = 9 \sec(\theta)$.

Step 2: Substituting into the integral

Now substitute these expressions back into the integral: $I = \int \frac{9 \sec(\theta)}{9 \tan(\theta)} \cdot 9 \sec^2(\theta) \, d\theta$ Simplifying: $I = \int \frac{\sec(\theta)}{\tan(\theta)} \cdot 9 \sec^2(\theta) \, d\theta$ $I = 9 \int \frac{\sec^3(\theta)}{\tan(\theta)} \, d\theta$

Step 3: Simplify the integrand

We can rewrite $\frac{\sec^3(\theta)}{\tan(\theta)}$ as: $\frac{\sec^3(\theta)}{\tan(\theta)} = \frac{1}{\sin(\theta)} \quad \text{(since } \sec(\theta) = \frac{1}{\cos(\theta)} \text{ and } \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)})$ Thus, the integral becomes: $I = 9 \int \frac{d\theta}{\sin(\theta)}$

This is a standard integral, where: $\int \frac{1}{\sin(\theta)} \, d\theta = \ln \left| \tan\left(\frac{\theta}{2}\right) \right| + C$

Step 4: Back-substitute $\theta$

Now we substitute back $\theta$ in terms of $x$. Recall that $x = 9 \tan(\theta)$, so: $\tan(\theta) = \frac{x}{9}$ Thus, the integral becomes: $I = 9 \ln\left| \frac{x}{9} + \sqrt{1 + \frac{x^2}{81}} \right| + C$

Final Answer:

$I = 9 \ln\left| x + \sqrt{x^2 + 81} \right| + C$

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Would you like further details on any step?

Here are 5 related questions:

1. How do trigonometric substitutions work in integrals?
2. What is the Pythagorean identity in trigonometry?
3. How do you compute the integral of secant functions?
4. What are other methods to solve integrals involving square roots?
5. Can the integral of secant be solved using different approaches?

Tip: When dealing with integrals involving square roots of sums of squares, trigonometric substitutions are often a powerful tool!