JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 21)
If $$\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C} $$
where C is a constant of integration, then the function ƒ(x) is equal to
where C is a constant of integration, then the function ƒ(x) is equal to
$${3 \over {{x^2}}}$$
$$ - {1 \over {6{x^3}}}$$
$$ - {1 \over {2{x^3}}}$$
$$ - {1 \over {2{x^2}}}$$
Explanation
I = $$\int {{{dx} \over {{x^3}{{\left( {1 + {x^6}} \right)}^{{2 \over 3}}}}}} $$
= $$\int {{{dx} \over {{x^7}{{\left( {{1 \over {{x^6}}} + 1} \right)}^{{2 \over 3}}}}}} $$
Let $${{1 \over {{x^6}}} + 1}$$ = t
$$ \Rightarrow $$ $${{ - 6} \over {{x^7}}}dx = dt$$
$$ \Rightarrow $$ $${{dx} \over {{x^7}}} = - {{dt} \over 6}$$
$$ \therefore $$ I = $$ - {1 \over 6}\int {{{dt} \over {{t^{{2 \over 3}}}}}} $$
= $$ - {1 \over 6}\left[ {{{{t^{{1 \over 3}}}} \over {{1 \over 3}}}} \right] + C$$
= $$ - {1 \over 2}{\left[ {{1 \over {{x^6}}} + 1} \right]^{{1 \over 3}}} + C$$
= $$ - {1 \over 2}{{{{\left( {1 + {x^6}} \right)}^{{1 \over 3}}}} \over {{x^2}}} \times {x \over x} + C$$
= $$x.\left( { - {1 \over {2{x^3}}}} \right).{\left( {1 + {x^6}} \right)^{{1 \over 3}}} + C$$
$$ \therefore $$ f(x) = $${ - {1 \over {2{x^3}}}}$$
= $$\int {{{dx} \over {{x^7}{{\left( {{1 \over {{x^6}}} + 1} \right)}^{{2 \over 3}}}}}} $$
Let $${{1 \over {{x^6}}} + 1}$$ = t
$$ \Rightarrow $$ $${{ - 6} \over {{x^7}}}dx = dt$$
$$ \Rightarrow $$ $${{dx} \over {{x^7}}} = - {{dt} \over 6}$$
$$ \therefore $$ I = $$ - {1 \over 6}\int {{{dt} \over {{t^{{2 \over 3}}}}}} $$
= $$ - {1 \over 6}\left[ {{{{t^{{1 \over 3}}}} \over {{1 \over 3}}}} \right] + C$$
= $$ - {1 \over 2}{\left[ {{1 \over {{x^6}}} + 1} \right]^{{1 \over 3}}} + C$$
= $$ - {1 \over 2}{{{{\left( {1 + {x^6}} \right)}^{{1 \over 3}}}} \over {{x^2}}} \times {x \over x} + C$$
= $$x.\left( { - {1 \over {2{x^3}}}} \right).{\left( {1 + {x^6}} \right)^{{1 \over 3}}} + C$$
$$ \therefore $$ f(x) = $${ - {1 \over {2{x^3}}}}$$
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