The correct option is
B arcsin(x+2)5+CLet I=∫dx√21−4x−x2
∫cscucotu du=−cscu+C∫du√a2−u2=arcsin ua+C
∫dua2+u2=1aarctan ua+C
∫duu√u2−a2=1aarcsecua+C
∫duu2−a2=12aln∣∣∣u−au+a∣∣∣+C
∫dua2−u2=12aln∣∣∣a+ua−u∣∣∣+CTaking only the denominator part, split 21 as 25−4
Now this 4 goes with the other terms to make another term u2, where u=x+2.
⇒I=∫dx√52−(x+2)2=arcsin (x+2)5+C