If f3x−43x+4=x+2, then ∫f(x)dx is
ex+2log3x−43x+4+c
−83log|1−x|+23x+c
83log|1−x|+x3+c
e[(3x−4)(3x+4)]−x22−2x+c
Put x=3t−43t+4=1−83t+4,dx=24(3t+4)2dt
Now ∫f(x)dx=∫f3t−43t+424(3t+4)2dt=∫t+2(3t+4)2(24)dt=8∫3t+4+2(3t+4)2dt
=8∫13t+4+2(3t+4)2dt=83log|3t+4|−2383t+4+C1=83log81−x−23(1−x)+C1=−83log|1−x|+23x+C