Key Concepts and Formulas

• Substitution Method for Integration: If we have an integral of the form $\int f(g(x))g'(x) dx$, we can substitute $u = g(x)$, which implies $du = g'(x) dx$. The integral then becomes $\int f(u) du$.
• Integral of $\frac{1}{x}$: $\int \frac{1}{x} dx = \ln|x| + C$
• Algebraic manipulation to simplify expressions.

Step-by-Step Solution

Step 1: Find an expression for x in terms of t

Let $t = \frac{x - 4}{x + 2}$. Our goal is to find $x$ in terms of $t$ so we can substitute and find an expression for $f(t)$. $t = \frac{x - 4}{x + 2}$ $t(x + 2) = x - 4$ $tx + 2t = x - 4$ $tx - x = -4 - 2t$ $x(t - 1) = -2t - 4$ $x = \frac{-2t - 4}{t - 1}$ $x = \frac{2t + 4}{1 - t}$ $x = \frac{2(t + 2)}{1 - t}$

Step 2: Substitute the expression for x into the given functional equation

We are given $f\left(\frac{x - 4}{x + 2}\right) = 2x + 1$. Since we let $t = \frac{x - 4}{x + 2}$, we have $f(t) = 2x + 1$. Now, substitute $x = \frac{2(t + 2)}{1 - t}$ into the equation: $f(t) = 2\left(\frac{2(t + 2)}{1 - t}\right) + 1$ $f(t) = \frac{4(t + 2)}{1 - t} + 1$ $f(t) = \frac{4t + 8}{1 - t} + 1$

Step 3: Simplify the expression for f(t)

Combine the terms to get a simplified expression for $f(t)$. $f(t) = \frac{4t + 8 + (1 - t)}{1 - t}$ $f(t) = \frac{3t + 9}{1 - t}$

Step 4: Rewrite the expression to facilitate integration

Rewrite the numerator so that we can separate the fraction into simpler terms. $f(t) = \frac{3t + 9}{1 - t} = \frac{-3(1 - t) + 12}{1 - t}$ $f(t) = \frac{-3(1 - t)}{1 - t} + \frac{12}{1 - t}$ $f(t) = -3 + \frac{12}{1 - t}$

Step 5: Find f(x)

Replace $t$ with $x$ to obtain $f(x)$. $f(x) = -3 + \frac{12}{1 - x}$

Step 6: Integrate f(x)

Integrate $f(x)$ with respect to $x$. $\int f(x) dx = \int \left(-3 + \frac{12}{1 - x}\right) dx$ $\int f(x) dx = \int -3 dx + \int \frac{12}{1 - x} dx$ $\int f(x) dx = -3x + 12 \int \frac{1}{1 - x} dx$ Let $u = 1 - x$, then $du = -dx$, so $dx = -du$. $\int f(x) dx = -3x + 12 \int \frac{1}{u} (-du)$ $\int f(x) dx = -3x - 12 \int \frac{1}{u} du$ $\int f(x) dx = -3x - 12 \ln|u| + C$ $\int f(x) dx = -3x - 12 \ln|1 - x| + C$ $\int f(x) dx = -12 \ln|1 - x| - 3x + C$

Common Mistakes & Tips

• Sign errors: Be careful with signs, especially when substituting and simplifying. The derivative of $(1-x)$ is $-1$, which affects the integration.
• Absolute value: Remember to include the absolute value when integrating $\frac{1}{x}$, as the logarithm is only defined for positive arguments.
• Algebraic manipulations: Take your time with algebraic steps to avoid errors.

Summary

We found an expression for $x$ in terms of $t$ by manipulating the equation $t = \frac{x - 4}{x + 2}$. We substituted this into the given functional equation to find $f(t)$. Then we replaced $t$ with $x$ to get $f(x)$. Finally, we integrated $f(x)$ to get the desired result: $-12 \ln|1 - x| - 3x + C$.

Final Answer

The final answer is \boxed{-12 \log_e |1 - x| - 3x + C}, which corresponds to option (B).

However, the "Correct Answer" given is (A) $12 \log_e |1 - x| + 3x + C$. Let's re-examine the solution. The error is in the provided answer. It should be option B.

The final answer is \boxed{-12 \log_e |1 - x| - 3x + C}, which corresponds to option (B).