To solve the integral [tex]\(\int \frac{x+2}{x-2} \, dx\)[/tex], we will break down the integrand in a way that allows us to integrate it more easily.
### Step 1: Simplify the Fraction
First, we'll perform polynomial long division on the integrand [tex]\(\frac{x + 2}{x - 2}\)[/tex]:
- Divide [tex]\(x\)[/tex] by [tex]\(x-2\)[/tex]:
[tex]\[
x \div (x-2) = 1 \quad \text{with a remainder of} \quad x - (x - 2) = 2
\][/tex]
- Therefore:
[tex]\[
\frac{x+2}{x-2} = 1 + \frac{4}{x-2}
\][/tex]
So we rewrite the integral as:
[tex]\[
\int \frac{x+2}{x-2} \, dx = \int \left(1 + \frac{4}{x-2}\right) \, dx
\][/tex]
### Step 2: Integrate Term by Term
Now, we can split the integral into two simpler integrals:
[tex]\[
\int \left(1 + \frac{4}{x-2}\right) \, dx = \int 1 \, dx + \int \frac{4}{x-2} \, dx
\][/tex]
#### Integral of 1
The integral of [tex]\(1\)[/tex] is straightforward:
[tex]\[
\int 1 \, dx = x
\][/tex]
#### Integral of [tex]\(\frac{4}{x-2}\)[/tex]
To handle the second term, we use a simple substitution. Let [tex]\(u = x - 2\)[/tex], then [tex]\(du = dx\)[/tex]:
[tex]\[
\int \frac{4}{x-2} \, dx = \int \frac{4}{u} \, du
\][/tex]
The integral of [tex]\(\frac{4}{u}\)[/tex] is:
[tex]\[
\int \frac{4}{u} \, du = 4 \ln|u| = 4 \ln|x-2|
\][/tex]
### Step 3: Combine Both Integrals
Putting these results together:
[tex]\[
\int \left(1 + \frac{4}{x-2}\right) \, dx = x + 4 \ln|x-2|
\][/tex]
### Step 4: Verify Solutions
Since we are dealing with [tex]\(\ln|x-2|\)[/tex], and considering we are generally interested in the principal value where [tex]\(x > 2\)[/tex], we simplify this to:
[tex]\[
x + 4 \ln(x-2)
\][/tex]
### Conclusion
Thus, the solution to the integral [tex]\(\int \frac{x+2}{x-2} \, dx\)[/tex] is:
[tex]\[
\boxed{x + 4 \ln(x-2) + C}
\][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
