Sure! Let's solve the integral [tex]\(\int \frac{(x-3)}{(x+3)} \, dx\)[/tex] step by step.
Step 1: Begin with the original integral:
[tex]\[
\int \frac{(x-3)}{(x+3)} \, dx
\][/tex]
Step 2: Simplify the expression inside the integral by dividing the numerator by the denominator. We can decompose [tex]\(\frac{(x-3)}{(x+3)}\)[/tex] as follows:
[tex]\[
\frac{x-3}{x+3} = \frac{x+3-6}{x+3} = \frac{x+3}{x+3} - \frac{6}{x+3}
\][/tex]
Step 3: Split the integral into two simpler integrals:
[tex]\[
\int \left( \frac{x+3}{x+3} - \frac{6}{x+3} \right) \, dx = \int 1 \, dx - \int \frac{6}{x+3} \, dx
\][/tex]
Step 4: Solve the first integral:
[tex]\[
\int 1 \, dx = x
\][/tex]
Step 5: Solve the second integral. Let [tex]\(u = x + 3\)[/tex], then [tex]\(du = dx\)[/tex]:
[tex]\[
\int \frac{6}{x+3} \, dx = 6 \int \frac{1}{u} \, du = 6 \ln|u| = 6 \ln|x+3|
\][/tex]
Step 6: Put everything together:
[tex]\[
\int \frac{(x-3)}{(x+3)} \, dx = x - 6 \ln|x+3| + C
\][/tex]
Therefore, the integral [tex]\(\int \frac{(x-3)}{(x+3)} \, dx\)[/tex] evaluates to:
[tex]\[
x - 6 \ln|x+3| + C
\][/tex]
Where [tex]\(C\)[/tex] is the constant of integration.
