The integral ∫eeex·eex·exdx=...+C is equal to
eex
12e2eex
eeex
12eex
Explanation for the correct option:
∫eeex·eex·exdx
Let eeex=z
⇒eeex·eex·exdx=dz⇒∫dz=z+C=eeex+C
Hence, option (C) is correct.