Verified Solution • 2025–26 Session moderate

QJEEM19Ma249f: JEE Main 2019 Mathematics - Class 12 JEEM19Ma249f Solution: If   dx = A(x) + C, for a suitable chosen integer m…

If   ∫1−x2x4\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} dx = A(x) (1−x2)m{\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 :

Multiple Choice Options

  • A 127x6{1 \over {27{x^6}}}
  • B −127x9{{ - 1} \over {27{x^9}}}
  • C 19x4{1 \over {9{x^4}}}
  • D 13x3{1 \over {3{x^3}}}

Step-by-Step Verified Solution

∫1−x2x4\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} dx = A(x) (1−x2)m{\left( {\sqrt {1 - {x^2}} } \right)^m} + C

∫∣x∣1x2−1x4 dx,\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,

Put   1x2−1=t⇒dtdx=−2x3{1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}

Case-I    x≥0x \ge 0

−12∫t dt ⇒t3/23+C- {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C

⇒−13(1x2−1)3/2\Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}

⇒(1−x2)3−3x2+C\Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C

A(x)=−13x3  A(x) = - {1 \over {3{x^3}}}\,\, and m=3m = 3

(A(x))m=(−13x3)3=−127x9{(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}

Case-II   x≤0x \le 0

We get   (1−x2)3−3x3+C{{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C

A(x)=1−3x3,   m=3A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3

(A(x))m=−127x9{(A(x))^m} = {{ - 1} \over {27{x^9}}}

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