If ∫400dx2x+1=loga, then a is:
∫400dx2x+1
=[log(2x+1)2]400 [ ∵∫1ax+b=log(ax+b)a+C]
=12[log(2(40)+1)−log(2(0)+1)]
==12[log81−log1]
=12log81
=log8112
=log√81
=log9
∴∫400dx2x+1=log9
It is given that, ∫400dx2x+1=loga
Hence, Option A is correct.