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Step-by-Step Solution
Step 1: Understand the Given Information
We are given a function in a slightly disguised manner:
$f\left(\frac{x - 4}{x + 2}\right) = 2x + 1.$
We need to find the integral of $f(x)$ with respect to $x$, i.e., $\int f(x)\,dx.$
Step 2: Express $f(t)$ in Terms of $t$
Let
\[
t = \frac{x - 4}{x + 2}.
\]
Then the given condition is $f(t) = 2x + 1.$ We will solve for $x$ in terms of $t$ to rewrite $f(t)$ fully in terms of $t$.
Step 2.1: Solve for $x$ in Terms of $t$
\[
t = \frac{x - 4}{x + 2}
\quad \Longrightarrow \quad
x - 4 = t(x + 2).
\]
\[
x - 4 = tx + 2t
\quad \Longrightarrow \quad
x - tx = 4 + 2t
\quad \Longrightarrow \quad
x(1 - t) = 2(t + 2).
\]
\[
x = \frac{2(t + 2)}{1 - t}.
\]
Step 2.2: Substitute $x$ Back into $f(t)$
\[
f(t) = 2x + 1 = 2 \left(\frac{2(t + 2)}{1 - t}\right) + 1.
\]
\[
f(t) = \frac{4(t + 2)}{1 - t} + 1.
\]
Combine into a single fraction:
\[
f(t) = \frac{4(t + 2)}{1 - t} + \frac{1 - t}{1 - t}
= \frac{4(t + 2) + (1 - t)}{1 - t}.
\]
\[
f(t) = \frac{4t + 8 + 1 - t}{1 - t}
= \frac{3t + 9}{1 - t}
= \frac{3(t + 3)}{1 - t}.
\]
Step 3: Rewrite $f(t)$ in a More Convenient Form
Observe that:
\[
\frac{3(t + 3)}{1 - t}
= 3 \cdot \frac{(t + 3)}{1 - t}.
\]
We can manipulate this expression further to separate terms. One effective way is to split it into simpler parts. Through suitable algebraic manipulation, one finds:
\[
f(t) = -3 + \frac{12}{1 - t}.
\]
(You can verify this by bringing them to a common denominator and simplifying.)
Step 4: Replace $t$ by $x$ to Obtain $f(x)$
Our goal is $\int f(x)\,dx.$ Based on the above step, we have:
\[
f(x) = -3 + \frac{12}{1 - x}.
\]
Therefore,
\[
\int f(x)\,dx
= \int \left(-3 + \frac{12}{1 - x}\right) dx.
\]
Step 5: Integrate Term by Term
We break the integral into two simpler integrals:
\[
\int \left(-3 + \frac{12}{1 - x}\right) dx
= \int -3 \, dx + \int \frac{12}{1 - x} \, dx.
\]
Step 5.1: Integrate $-3$
\[
\int -3 \, dx = -3x.
\]
Step 5.2: Integrate $\frac{12}{1 - x}$
Let us rewrite $\frac{12}{1 - x}$ thoughtfully. We can set $u = 1 - x$, then $du = -dx$, or $dx = -du$. Another common approach is to recall the integral of $\frac{1}{a - x}$.
\[
\int \frac{12}{1 - x} \, dx
= 12 \int \frac{1}{1 - x} \, dx.
\]
We know that
\[
\int \frac{1}{1 - x} \, dx = -\ln|1 - x|.
\]
Hence,
\[
12 \int \frac{1}{1 - x} \, dx = 12 \cdot \bigl[-\ln|1 - x|\bigr]
= -12 \ln|1 - x|.
\]
Step 6: Combine the Results
\[
\int f(x)\,dx
= \int \left(-3 + \frac{12}{1 - x}\right) dx
= -3x - 12 \ln|1 - x| + C,
\]
where $C$ is the constant of integration.
Step 7: Final Answer
Thus, the integral is
\[
\boxed{-12 \ln\lvert 1 - x\rvert \;-\; 3x \;+\; C}.
\]
This matches the correct choice:
$-12 \log_{e}|1 - x| - 3x + C$.
