If ∫√1−x2x4dx=A(x)(√1−x2)+C, for a suitable chosen integer m and a function A(x), where C is a constant of integration then (A(x))m equals :

The correct option is B 𕒵27x9
∫√1−x2x4dx=A(x)(√1−x2)m+C
∫|x|√1x2𕒵x4dx
Put 1x2𕒵=t⇒dtdx=𕒶x3
Case 𕒵x≥0
𕒵2∫√tdt⇒−t323+C⇒𕒵3(1x2𕒵)32⇒(√1−x2)3𕒷x2+C
A(x)=𕒵3x3 and m=3
(A(x))m=(𕒵3x3)3=𕒵27x9
Case-II x≤0
We get √(1−x2)3𕒷x3+C
A(x)=1𕒷x3, m=3
(A(x))m=𕒵27x9