Integral Calculus: JEE Main 2019 Mathematics question

JEE Main 2019 · 11 Jan, Morning shift · Q4017

JEE Main 201911 Jan, Morning shiftMathematicsMediumIntegral Calculus369× since 2002

\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 :

JEE Main 2019 · 11 Jan, Morning shift · Q4017

JEE Main 201911 Jan, Morning shiftMathematicsMediumIntegral Calculus369× since 2002

\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 :

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