Verified Solution • 2025–26 Session moderate

QJEEM19Ma3e53: JEE Main 2019 Mathematics - Class 12 JEEM19Ma3e53 Solution: If   x 5 .e 4x 3 dx = e 4x 3 f(x) + C, where C is a…

If   ∫ \int \, x5.e −- 4x3 dx = 148{1 \over {48}} e −- 4x3 f(x) + C, where C is a constant of inegration, then f(x) is equal to -

Multiple Choice Options

  • A −- 2x3 −- 1
  • B −- 2x3 + 1
  • C 4x3 + 1
  • D −- 4x3 −- 1

Step-by-Step Verified Solution

∫x5.e−4x3 dx=148e−4x3f(x)+c\int {{x^5}} .{e^{ - 4{x^3}}}\,dx = {1 \over {48}}{e^{ - 4{x^3}}}f\left( x \right) + c

Put   x3=t{x^3} = t

3x2 dx=dt3{x^2}\,dx = dt

∫x3.e−4x3. x2dx\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx

13∫t.e−4tdt{1 \over 3}\int {t.{e^{ - 4t}}dt}

13[t.e−4t−4−∫e−4t−4dt]{1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \int {{{{e^{ - 4t}}} \over { - 4}}dt} } \right]

−e−4t48[4t+1]+c- {{{e^{ - 4t}}} \over {48}}\left[ {4t + 1} \right] + c

−e−4x348[4x3+1]+c{{ - {e^{ - 4{x^3}}}} \over {48}}\left[ {4{x^3} + 1} \right] + c

∴\therefore    f(x)=−1−4x3f(x) = - 1 - 4{x^3}

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