The correct option is
A log(ex+√e2x−1)−sec−1(ex)+c∫ex−1√e2x−1dx.
∫ex√e2x−1⋅dx−∫1√e2x−1⋅dx
dex√e2x−1−∫1√e2x−1dx.∫1√e2x−1dx.
log∣ex+√e2x−1∣−∫1√e2x−1dx
Let ex=t.
exdx=dt
dx=1tdt
log∣ex+√e2x−1∣−∫dtt√t2−1
log∣ex+√e2x−1∣−sec−1(t)+c
=log∣ex+√e2x−1∣−sec−1(ex)−c.