JEE Math Practice Question
If $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :
- A.$${1 \over {27{x^6}}}$$
- B.$${{ - 1} \over {27{x^9}}}$$
- C.$${1 \over {9{x^4}}}$$
- D.$${1 \over {3{x^3}}}$$
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