Verified Solution • 2025–26 Session moderate

QJEEM19Ma073b: JEE Main 2019 Mathematics - Class 12 JEEM19Ma073b Solution: The integral is equal to : (where C is a constant of integration)

The integral ∫3x13+2x11(2x4+3x2+1)4 dx\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} \,dx is equal to : (where C is a constant of integration)

Multiple Choice Options

  • A x126(2x4+3x2+1)3{{{x^{12}}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + CC
  • B x46(2x4+3x2+1)3+C{{{x^4}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C
  • C x12(2x4+3x2+1)3+C{{{x^{12}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C
  • D x4(2x4+3x2+1)3+C{{{x^4}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C

Step-by-Step Verified Solution

∫3x13+2x11(2x4+3x2+1)4dx\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} dx

∫(3x3+2x5)dx(2+3x2+1x4)4\int {{{\left( {{3 \over {{x^3}}} + {2 \over {{x^5}}}} \right)dx} \over {{{\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right)}^4}}}}

Let   (2+3x2+1x4)=t\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right) = t

−12∫dtt4=16t3+C- {1 \over 2}\int {{{dt} \over {{t^4}}}} = {1 \over {6{t^3}}} + C

⇒x126(2x4+3x2+1)3+C\Rightarrow {{{x^{12}}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C

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