JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 19)
The integral $$\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} \,dx$$ is equal to : (where C is a constant of integration)
$${{{x^{12}}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}}$$ + $$C$$
$${{{x^4}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C$$
$${{{x^{12}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C$$
$${{{x^4}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C$$
Explanation
$$\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} dx$$
$$\int {{{\left( {{3 \over {{x^3}}} + {2 \over {{x^5}}}} \right)dx} \over {{{\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right)}^4}}}} $$
Let $$\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right) = t$$
$$ - {1 \over 2}\int {{{dt} \over {{t^4}}}} = {1 \over {6{t^3}}} + C$$
$$ \Rightarrow {{{x^{12}}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C$$
$$\int {{{\left( {{3 \over {{x^3}}} + {2 \over {{x^5}}}} \right)dx} \over {{{\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right)}^4}}}} $$
Let $$\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right) = t$$
$$ - {1 \over 2}\int {{{dt} \over {{t^4}}}} = {1 \over {6{t^3}}} + C$$
$$ \Rightarrow {{{x^{12}}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C$$
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