To solve the integral [tex]\(\int \frac{x+2}{x-2} \, dx\)[/tex], follow these steps:
1. Separate the Integrand:
Rewrite the integrand [tex]\(\frac{x+2}{x-2}\)[/tex] as a sum of fractions. We can make it simpler by performing polynomial long division.
[tex]\[
\frac{x+2}{x-2} = 1 + \frac{4}{x-2}
\][/tex]
This is because:
[tex]\[
\frac{x+2}{x-2} = \frac{(x-2) + 4}{x-2} = \frac{x-2}{x-2} + \frac{4}{x-2} = 1 + \frac{4}{x-2}
\][/tex]
2. Split the Integral:
Now we can split the integral into two simpler integrals:
[tex]\[
\int \frac{x+2}{x-2} \, dx = \int 1 \, dx + \int \frac{4}{x-2} \, dx
\][/tex]
3. Integrate Each Term:
- The first integral is straightforward:
[tex]\[
\int 1 \, dx = x
\][/tex]
- The second integral involves a logarithmic function. To integrate [tex]\(\int \frac{4}{x-2} \, dx\)[/tex], recognize that this is a standard form [tex]\(\int \frac{1}{u} \, du\)[/tex] with [tex]\( u = x-2 \)[/tex]:
[tex]\[
\int \frac{4}{x-2} \, dx = 4 \int \frac{1}{x-2} \, dx
\][/tex]
Let [tex]\( u = x-2 \)[/tex], then [tex]\( du = dx \)[/tex]. Hence:
[tex]\[
4 \int \frac{1}{u} \, du = 4 \ln |u| + C = 4 \ln |x-2| + C
\][/tex]
4. Combine the Results:
Combining the results of the two integrals, we get:
[tex]\[
\int \frac{x+2}{x-2} \, dx = x + 4 \ln |x-2| + C
\][/tex]
Thus, the final answer to the integral is:
[tex]\[
x + 4 \ln |x-2| + C
\][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
