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Indefinite Integrals question

2019 · 11 Jan · Shift 1 · Q32

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PYQ Library
JEE Main
Mathematics
Indefinite Integrals
2019 · 11 Jan · Shift 1 · Q32

Indefinite Integrals question

2019 · 11 Jan · Shift 1 · Q32

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1

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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