To evaluate the integral [tex]\(\int \frac{3}{9 + x^2} \, dx\)[/tex], we can recognize it follows a standard form that involves the arctangent function. Let's go through the steps to solve this integral:
1. Recognize the Standard Integral Form:
The integral [tex]\(\int \frac{dx}{a^2 + x^2}\)[/tex] has a standard result:
[tex]\[
\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C
\][/tex]
where [tex]\(a\)[/tex] is a constant.
2. Rewrite the Integral:
In our case, we need to examine [tex]\(\int \frac{3 \, dx}{9 + x^2}\)[/tex]. Notice that the denominator can be expressed as [tex]\(3^2 + x^2\)[/tex].
3. Factor Out Constants:
Separate the constant factor in the numerator:
[tex]\[
\int \frac{3 \, dx}{9 + x^2} = 3 \int \frac{dx}{3^2 + x^2}
\][/tex]
4. Apply the Standard Result:
Using the known result for the standard form [tex]\(\frac{1}{a} \arctan\left(\frac{x}{a}\right) + C\)[/tex], with [tex]\(a = 3\)[/tex]:
[tex]\[
\int \frac{dx}{3^2 + x^2} = \frac{1}{3} \arctan\left(\frac{x}{3}\right) + C
\][/tex]
Now, multiply by the factor of 3 we factored out earlier:
[tex]\[
3 \int \frac{dx}{3^2 + x^2} = 3 \left( \frac{1}{3} \arctan\left(\frac{x}{3}\right) + C \right)
\][/tex]
Simplify:
[tex]\[
3 \cdot \frac{1}{3} \arctan\left(\frac{x}{3}\right) + 3C = \arctan\left(\frac{x}{3}\right) + C'
\][/tex]
Since [tex]\(C'\)[/tex] is just another constant (we can merge constants of integration), we denote it simply as [tex]\(C\)[/tex].
Therefore, the integral evaluates to:
[tex]\[
\int \frac{3}{9 + x^2} \, dx = \arctan\left(\frac{x}{3}\right) + C
\][/tex]
5. Select the Correct Answer:
Comparing this result with the given options:
- (A) [tex]\(\tan^{-1}\left(\frac{x}{3}\right) + C\)[/tex]
- (B) [tex]\(\sin^{-1}\left(\frac{x}{4}\right) + C\)[/tex]
- (C) [tex]\(\tan^{-1}\left(\frac{x}{4}\right) + C\)[/tex]
- (D) [tex]\(\sin^{-1}\left(\frac{x}{3}\right) + C\)[/tex]
The correct choice is:
[tex]\[ \boxed{\text{(A) } \tan^{-1}\left(\frac{x}{3}\right) + C} \][/tex]
