Verified Solution • 2025–26 Session moderate

QJEEM16Mabe36: JEE Main 2016 Mathematics - Class 12 JEEM16Mabe36 Solution: The integral is equal to :

The integral ∫2x12+5x9(x5+x3+1)3dx\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx is equal to :

Multiple Choice Options

  • A x52(x5+x3+1)2+C{{{x^5}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C
  • B −x102(x5+x3+1)2+C{{ - {x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C
  • C −x5(x5+x3+1)2+C{{{-x^5}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C
  • D x102(x5+x3+1)2+C{{ {x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C

Step-by-Step Verified Solution

∫2x12+5x9(x5+x3+1)3dx\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx

Dividing by x15{x^{15}} in numerator and denominator

∫2x3+5x6dx(1+1x2+15)3\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over 5}} \right)}^3}}}}

Substitute 1+1x2+1x5=t1 + {1 \over {{x^2}}} + {1 \over {{x^5}}} = t

⇒(−2x3−5x6)dx=dt\Rightarrow \left( {{{ - 2} \over {{x^3}}} - {5 \over {{x^6}}}} \right)dx = dt

⇒(2x3+5x6)dx=−dt\Rightarrow \left( {{2 \over {{x^3}}} + {5 \over {{x^6}}}} \right)dx = - dt

This gives,

∫2x3+5x6dx(1+1x2+1x5)3\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^3}}}}

=∫−dtt3=12t2+C= \int {{{ - dt} \over {{t^3}}}} = {1 \over {2{t^2}}} + C

=12(1+1x2+1x5)2+C= {1 \over {2{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^2}}} + C

=x102(x5+x3+1)2+C= {{{x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C

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