JEE MAIN - Mathematics (2016 (Offline) - No. 8)
The integral $$\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx$$ is equal to :
$${{{x^5}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$$
$${{ - {x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$$
$${{{-x^5}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$$
$${{ {x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$$
Explanation
$$\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx$$
Dividing by $${x^{15}}$$ in numerator and denominator
$$\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over 5}} \right)}^3}}}} $$
Substitute $$1 + {1 \over {{x^2}}} + {1 \over {{x^5}}} = t$$
$$ \Rightarrow \left( {{{ - 2} \over {{x^3}}} - {5 \over {{x^6}}}} \right)dx = dt$$
$$ \Rightarrow \left( {{2 \over {{x^3}}} + {5 \over {{x^6}}}} \right)dx = - dt$$
This gives,
$$\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^3}}}} $$
$$ = \int {{{ - dt} \over {{t^3}}}} = {1 \over {2{t^2}}} + C$$
$$ = {1 \over {2{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^2}}} + C$$
$$ = {{{x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$$
Dividing by $${x^{15}}$$ in numerator and denominator
$$\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over 5}} \right)}^3}}}} $$
Substitute $$1 + {1 \over {{x^2}}} + {1 \over {{x^5}}} = t$$
$$ \Rightarrow \left( {{{ - 2} \over {{x^3}}} - {5 \over {{x^6}}}} \right)dx = dt$$
$$ \Rightarrow \left( {{2 \over {{x^3}}} + {5 \over {{x^6}}}} \right)dx = - dt$$
This gives,
$$\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^3}}}} $$
$$ = \int {{{ - dt} \over {{t^3}}}} = {1 \over {2{t^2}}} + C$$
$$ = {1 \over {2{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^2}}} + C$$
$$ = {{{x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$$
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