JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 21)
If $$f(x) = \int {{{5{x^8} + 7{x^6}} \over {{{({x^2} + 1 + 2{x^7})}^2}}}dx,(x \ge 0),f(0) = 0} $$ and $$f(1) = {1 \over K}$$, then the value of K is
Answer
4
Explanation
$$\int {{{5{x^8} + 7{x^6}} \over {{{(2{x^7} + {x^2} + 1)}^2}}}dx = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}} \right)}^2}}}dx} } $$
$$\int {{{{5 \over {{x^6}}} + {7 \over {{x^8}}}} \over {{{\left( {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}} \right)}^2}}}dx} $$
put $$2 + {1 \over {{x^5}}} + {1 \over {{x^7}}} = t$$
$$ \Rightarrow - \left( {{5 \over {{x^6}}} + {7 \over {{x^8}}}} \right)dx = dt$$
$$\int {{{ - dt} \over {{t^2}}} = {1 \over t} + c} $$
$$ \Rightarrow f(x) = {1 \over {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}}} + C = {{{x^7}} \over {2{x^7} + 1 + {x^2}}} + C$$
$$f(0) = 0 \Rightarrow C = 0$$
$$f(x) = {1 \over 4} = {1 \over k}$$
$$ \Rightarrow k = 4$$
$$\int {{{{5 \over {{x^6}}} + {7 \over {{x^8}}}} \over {{{\left( {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}} \right)}^2}}}dx} $$
put $$2 + {1 \over {{x^5}}} + {1 \over {{x^7}}} = t$$
$$ \Rightarrow - \left( {{5 \over {{x^6}}} + {7 \over {{x^8}}}} \right)dx = dt$$
$$\int {{{ - dt} \over {{t^2}}} = {1 \over t} + c} $$
$$ \Rightarrow f(x) = {1 \over {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}}} + C = {{{x^7}} \over {2{x^7} + 1 + {x^2}}} + C$$
$$f(0) = 0 \Rightarrow C = 0$$
$$f(x) = {1 \over 4} = {1 \over k}$$
$$ \Rightarrow k = 4$$
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