Key Concepts and Formulas

• Substitution Method for Integration: If $\int f(g(x))g'(x) \, dx$ is given, substitute $u = g(x)$, so $du = g'(x) \, dx$. This simplifies the integral to $\int f(u) \, du$.
• Algebraic Manipulation: Rearranging equations to solve for a variable.
• Basic Integration Formulas: $\int \frac{1}{x} \, dx = \log|x| + C$ and $\int 1 \, dx = x + C$.

Step-by-Step Solution

Step 1: Find the expression for $f(x)$. We are given $f\left(\frac{3x - 4}{3x + 4}\right) = x + 2$. Let $t = \frac{3x - 4}{3x + 4}$. Our goal is to express $x$ in terms of $t$. $t = \frac{3x - 4}{3x + 4}$ $t(3x + 4) = 3x - 4$ $3tx + 4t = 3x - 4$ $3tx - 3x = -4 - 4t$ $x(3t - 3) = -4 - 4t$ $x = \frac{-4 - 4t}{3t - 3} = \frac{4t + 4}{3 - 3t}$ Now, substitute this expression for $x$ into the equation $f(t) = x + 2$: $f(t) = \frac{4t + 4}{3 - 3t} + 2$ $f(t) = \frac{4t + 4 + 2(3 - 3t)}{3 - 3t} = \frac{4t + 4 + 6 - 6t}{3 - 3t} = \frac{10 - 2t}{3 - 3t}$ Therefore, $f(x) = \frac{10 - 2x}{3 - 3x} = \frac{2x - 10}{3x - 3}$.

Step 2: Evaluate the integral $\int f(x) \, dx$. We need to find $\int \frac{2x - 10}{3x - 3} \, dx$. $\int \frac{2x - 10}{3x - 3} \, dx = \frac{1}{3} \int \frac{2x - 10}{x - 1} \, dx$ We can rewrite the numerator as $2x - 10 = 2(x - 1) - 8$. $\frac{1}{3} \int \frac{2(x - 1) - 8}{x - 1} \, dx = \frac{1}{3} \int \left(2 - \frac{8}{x - 1}\right) \, dx$ $= \frac{1}{3} \left( \int 2 \, dx - \int \frac{8}{x - 1} \, dx \right) = \frac{1}{3} \left( 2x - 8 \int \frac{1}{x - 1} \, dx \right)$ $= \frac{1}{3} (2x - 8 \log|x - 1|) + C = \frac{2}{3}x - \frac{8}{3} \log|x - 1| + C$

Step 3: Compare the result with the given form and find A and B. We are given that $\int f(x) \, dx = A \log|1 - x| + Bx + C$. Our result is $\int f(x) \, dx = \frac{2}{3}x - \frac{8}{3} \log|x - 1| + C$. Since $\log|x - 1| = \log|-(1 - x)| = \log|-1| + \log|1 - x| = \log|1 - x|$, we can rewrite our result as: $\int f(x) \, dx = -\frac{8}{3} \log|1 - x| + \frac{2}{3}x + C$ Comparing this with $A \log|1 - x| + Bx + C$, we have $A = -\frac{8}{3}$ and $B = \frac{2}{3}$.

Step 4: Write the ordered pair (A, B). The ordered pair (A, B) is $\left(-\frac{8}{3}, \frac{2}{3}\right)$.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when manipulating equations and integrating. A small sign error can lead to an incorrect answer.
• Constant of Integration: Don't forget to add the constant of integration, C, after performing an indefinite integral.
• Logarithm Properties: Remember that $\log|x - 1| = \log|1 - x|$ because $\log|-1| = 0$ when considering the absolute value.

Summary

We first found an explicit expression for $f(x)$ by substituting $t = \frac{3x - 4}{3x + 4}$ and solving for $x$ in terms of $t$. Then we substituted this expression into the equation $f(t) = x + 2$ to obtain $f(x)$. Next, we evaluated the indefinite integral of $f(x)$ using algebraic manipulation and basic integration formulas. Finally, we compared our result with the given form $A \log|1 - x| + Bx + C$ to determine the values of A and B, which are $A = -\frac{8}{3}$ and $B = \frac{2}{3}$.

The final answer is \boxed{\left( { - {8 \over 3},{2 \over 3}} \right)}, which corresponds to option (B).