JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 13)
If $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :
$${1 \over {27{x^6}}}$$
$${{ - 1} \over {27{x^9}}}$$
$${1 \over {9{x^4}}}$$
$${1 \over {3{x^3}}}$$
Explanation
$$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C
$$\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,$$
Put $${1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}$$
Case-I $$x \ge 0$$
$$ - {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C$$
$$ \Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}$$
$$ \Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C$$
$$A(x) = - {1 \over {3{x^3}}}\,\,$$ and $$m = 3$$
$${(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}$$
Case-II $$x \le 0$$
We get $${{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C$$
$$A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3$$
$${(A(x))^m} = {{ - 1} \over {27{x^9}}}$$
$$\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,$$
Put $${1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}$$
Case-I $$x \ge 0$$
$$ - {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C$$
$$ \Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}$$
$$ \Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C$$
$$A(x) = - {1 \over {3{x^3}}}\,\,$$ and $$m = 3$$
$${(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}$$
Case-II $$x \le 0$$
We get $${{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C$$
$$A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3$$
$${(A(x))^m} = {{ - 1} \over {27{x^9}}}$$
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