To find the integral of the function $\frac{2}{9 + x^2}$, we can use a standard integration formula involving inverse trigonometric functions.

Integral Formula

The integral we need is:

$\int \frac{2}{9 + x^2} \, dx$

This is of the form $\int \frac{a}{a^2 + x^2} \, dx$, which has a standard result:

$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$

Applying the Formula

In our case, $a^2 = 9$, so $a = 3$. The integral can be rewritten as:

$2 \int \frac{1}{9 + x^2} \, dx = 2 \cdot \frac{1}{3} \arctan\left(\frac{x}{3}\right) + C$

Simplifying gives:

$\frac{2}{3} \arctan\left(\frac{x}{3}\right) + C$

Final Answer

Thus, the integral of $\frac{2}{9 + x^2}$ is:

$\int \frac{2}{9 + x^2} \, dx = \frac{2}{3} \arctan\left(\frac{x}{3}\right) + C$

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Related Questions:

1. How do you integrate $\frac{1}{x^2 + a^2}$ in general?
2. What are the common integral forms involving inverse trigonometric functions?
3. Can you find the derivative of $\arctan(x)$?
4. How do we evaluate definite integrals involving $\arctan(x)$?
5. What is the importance of using trigonometric substitutions in integration?

Tip: Remember that recognizing patterns and forms is key to applying standard integration formulas correctly.