Verified Solution • 2025–26 Session moderate

QJEEM19Ma9494: JEE Main 2019 Mathematics - Class 12 JEEM19Ma9494 Solution: If where C is a constant of integration, then the function ƒ(x) …

If ∫dxx3(1+x6)2/3=xf(x)(1+x6)13+C\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C}
where C is a constant of integration, then the function ƒ(x) is equal to

Multiple Choice Options

  • A 3x2{3 \over {{x^2}}}
  • B −16x3- {1 \over {6{x^3}}}
  • C −12x3- {1 \over {2{x^3}}}
  • D −12x2- {1 \over {2{x^2}}}

Step-by-Step Verified Solution

I = ∫dxx3(1+x6)23\int {{{dx} \over {{x^3}{{\left( {1 + {x^6}} \right)}^{{2 \over 3}}}}}}

= ∫dxx7(1x6+1)23\int {{{dx} \over {{x^7}{{\left( {{1 \over {{x^6}}} + 1} \right)}^{{2 \over 3}}}}}}

Let 1x6+1{{1 \over {{x^6}}} + 1} = t

⇒\Rightarrow −6x7dx=dt{{ - 6} \over {{x^7}}}dx = dt

⇒\Rightarrow dxx7=−dt6{{dx} \over {{x^7}}} = - {{dt} \over 6}

∴\therefore I = −16∫dtt23- {1 \over 6}\int {{{dt} \over {{t^{{2 \over 3}}}}}}

= −16[t1313]+C- {1 \over 6}\left[ {{{{t^{{1 \over 3}}}} \over {{1 \over 3}}}} \right] + C

= −12[1x6+1]13+C- {1 \over 2}{\left[ {{1 \over {{x^6}}} + 1} \right]^{{1 \over 3}}} + C

= −12(1+x6)13x2×xx+C- {1 \over 2}{{{{\left( {1 + {x^6}} \right)}^{{1 \over 3}}}} \over {{x^2}}} \times {x \over x} + C

= x.(−12x3).(1+x6)13+Cx.\left( { - {1 \over {2{x^3}}}} \right).{\left( {1 + {x^6}} \right)^{{1 \over 3}}} + C

∴\therefore f(x) = −12x3{ - {1 \over {2{x^3}}}}

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