If f(3x/(3x+4)) = x+2, x ≠ -4/3, and ∫f(x)dx = Alog|1-x| + Bx + C, then the ordered pair (A,B) is equal to (where C is a constant of integration)

∫f(x)dx = Alog|1−x| + Bx + C
Differentiate w.r.t x
f(x) = −A/(1−x) + B
f(3x/(3x+4)) = −A/(1−(3x/(3x+4))) + B = −A/((3x+4−3x)/(3x+4)) + B = −A(3x+4)/4 + B
→ x + 2 = −A(3x+4)/4 + B
→ x + 2 = (−3Ax)/4 − A + B
Comparing coefficients of x:
1 = −3A/4 → A = −4/3
Comparing constant terms:
2 = −A + B → 2 = 4/3 + B → B = 2 − 4/3 = 2/3
Hence, (A,B) = (−4/3, 2/3)