If $$f\left(\frac{3x - 4}{3x + 4}\right) = x + 2$$, $$x \neq -\frac{4}{3}$$, and $$\int f(x)dx = A \log|1 - x| + Bx + C$$, then the ordered pair $$(A, B)$$ is equal to

The relation is given as $$f\!\left(\dfrac{3x-4}{\,3x+4\,}\right)=x+2,\qquad x\neq-\dfrac43.$$

Put $$y=\dfrac{3x-4}{\,3x+4\,}\,.$$ Then

$$y(3x+4)=3x-4\;-(1)$$

Expanding the left side of $$-(1)$$ gives $$3xy+4y=3x-4\;-(2)$$

Bring every term containing $$x$$ to the left and the constants to the right:

$$3xy-3x=-4-4y\;-(3)$$

Factor $$x$$ on the left side of $$-(3)$$:

$$3x(y-1)=-4(1+y)\;-(4)$$

Solve $$-(4)$$ for $$x$$:

$$x=\dfrac{-4(1+y)}{3(y-1)}=\dfrac{4(1+y)}{3(1-y)}\;-(5)$$

According to the original relation, $$f(y)=x+2.$$ Substituting the value of $$x$$ from $$-(5)$$, we get

$$f(y)=\dfrac{4(1+y)}{3(1-y)}+2\;-(6)$$

Write the constant $$2$$ with the same denominator $$3(1-y)$$ to combine the fractions:

$$2=\dfrac{2\cdot3(1-y)}{3(1-y)}=\dfrac{6(1-y)}{3(1-y)}\;-(7)$$

Add $$-(6)$$ and $$-(7)$$:

$$f(y)=\dfrac{4(1+y)+6(1-y)}{3(1-y)}\;-(8)$$

Simplify the numerator of $$-(8)$$:

$$4(1+y)=4+4y,\quad 6(1-y)=6-6y,$$

so $$4+4y+6-6y=10-2y\;-(9)$$

Therefore

$$f(y)=\dfrac{10-2y}{3(1-y)}=\dfrac{2(5-y)}{3(1-y)}\;-(10)$$

Replace the dummy variable $$y$$ by $$x$$ to obtain the required expression of the function:

$$f(x)=\dfrac{2(5-x)}{3(1-x)}\;-(11)$$

The integral to be evaluated is

$$\int f(x)\,dx=\int\dfrac{2(5-x)}{3(1-x)}\,dx\;-(12)$$

Pull the constant $$\dfrac23$$ outside the integral:

$$\int f(x)\,dx=\dfrac23\int\dfrac{5-x}{1-x}\,dx\;-(13)$$

The integrand in $$-(13)$$ can be decomposed. Observe that

$$\dfrac{5-x}{1-x}=1+\dfrac{4}{1-x}\;-(14)$$

Hence $$-(13)$$ becomes

$$\int f(x)\,dx=\dfrac23\left(\int1\,dx+4\int\dfrac{dx}{1-x}\right)\;-(15)$$

Compute the two elementary integrals:

$$\int1\,dx=x\;-(16)$$

For the second integral, use the substitution $$u=1-x\;\Rightarrow\;du=-dx,$$ so

$$4\int\dfrac{dx}{1-x}=4\int\dfrac{-du}{u}=-4\ln|u|=-4\ln|1-x|\;-(17)$$

Insert $$-(16)$$ and $$-(17)$$ into $$-(15)$$:

$$\int f(x)\,dx=\dfrac23\left(x-4\ln|1-x|\right)+C\;-(18)$$

Distribute the factor $$\dfrac23$$ in $$-(18)$$:

$$\int f(x)\,dx=\dfrac23\,x-\dfrac{8}{3}\,\ln|1-x|+C\;-(19)$$

The problem states that the antiderivative has the form $$A\log|1-x|+Bx+C.$$ Comparing this with $$-(19)$$ gives

$$A=-\dfrac{8}{3},\qquad B=\dfrac{2}{3}\;-(20)$$

Thus the ordered pair $$(A,B)$$ equals $$\left(-\dfrac{8}{3},\dfrac{2}{3}\right).$$

Hence, the correct answer is Option B.