First, let's consider the integral [tex]\(\int \frac{12}{x^2 - 9} \, dx\)[/tex].
To solve this, we can use the method of partial fraction decomposition.
First, we recognize that [tex]\(x^2 - 9\)[/tex] can be factored as follows:
[tex]\[ x^2 - 9 = (x - 3)(x + 3) \][/tex]
Now we can decompose the fraction [tex]\(\frac{12}{x^2 - 9}\)[/tex] into partial fractions:
[tex]\[ \frac{12}{(x - 3)(x + 3)} = \frac{A}{x - 3} + \frac{B}{x + 3} \][/tex]
To determine the values of [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can set up the equation:
[tex]\[ 12 = A(x + 3) + B(x - 3) \][/tex]
We get:
[tex]\[ 12 = A(x + 3) + B(x - 3) \][/tex]
We need to solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex] by substituting convenient values for [tex]\(x\)[/tex]:
1. Let [tex]\(x = 3\)[/tex]:
[tex]\[ 12 = A(3 + 3) + B(3 - 3) \][/tex]
[tex]\[ 12 = 6A \][/tex]
[tex]\[ A = 2 \][/tex]
2. Let [tex]\(x = -3\)[/tex]:
[tex]\[ 12 = A(-3 + 3) + B(-3 - 3) \][/tex]
[tex]\[ 12 = -6B \][/tex]
[tex]\[ B = -2 \][/tex]
Thus, we have the partial fractions:
[tex]\[ \frac{12}{(x - 3)(x + 3)} = \frac{2}{x - 3} - \frac{2}{x + 3} \][/tex]
Now, we can integrate each term separately:
[tex]\[ \int \frac{12}{x^2 - 9} \, dx = \int \left( \frac{2}{x - 3} - \frac{2}{x + 3} \right) dx \][/tex]
Integrate each term:
[tex]\[ \int \frac{2}{x - 3} \, dx - \int \frac{2}{x + 3} \, dx \][/tex]
The integral of [tex]\(\frac{1}{x - 3}\)[/tex] is [tex]\(\ln|x - 3|\)[/tex], and the integral of [tex]\(\frac{1}{x + 3}\)[/tex] is [tex]\(\ln|x + 3|\)[/tex]. Thus:
[tex]\[ \int \frac{2}{x - 3} \, dx = 2 \ln|x - 3| \][/tex]
[tex]\[ \int \frac{2}{x + 3} \, dx = 2 \ln|x + 3| \][/tex]
So, the integral becomes:
[tex]\[ 2 \ln|x - 3| - 2 \ln|x + 3| + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
Given this, we need to multiply the integral result by 1.2:
[tex]\[ 1.2 \left(2 \ln|x - 3| - 2 \ln|x + 3|\right) = 2.4 \ln|x - 3| - 2.4 \ln|x + 3| + C \][/tex]
Therefore, the final answer is:
[tex]\[ \boxed{2.4 \ln|x - 3| - 2.4 \ln|x + 3| + C} \][/tex]
