JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 13)
The integral $$\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}} $$ is equal to :
(where C is a constant of integration)
(where C is a constant of integration)
$${1 \over 2}{\left( {{{x - 3} \over {x + 4}}} \right)^{{3 \over 7}}} + C$$
$${\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}} + C$$
$$ - {1 \over {13}}{\left( {{{x - 3} \over {x + 4}}} \right)^{{{13} \over 7}}} + C$$
-$${\left( {{{x - 3} \over {x + 4}}} \right)^{-{1 \over 7}}} + C$$
Explanation
$$\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}} $$
= $$\int {{{dx} \over {{{\left( {x + 4} \right)}^2}{{\left( {{{x - 3} \over {x + 4}}} \right)}^{{6 \over 7}}}}}} $$
Put $${{{x - 3} \over {x + 4}}}$$ = t
$$ \Rightarrow $$ $$\left\{ {{{\left( {x + 4} \right) - \left( {x - 3} \right)} \over {{{\left( {x + 4} \right)}^2}}}} \right\}dx$$ = dt
$$ \Rightarrow $$ $${{dx} \over {{{\left( {x + 4} \right)}^2}}} = {{dt} \over 7}$$
= $${1 \over 7}\int {{{dt} \over {{{\left( t \right)}^{{6 \over 7}}}}}} $$
= $${1 \over 7}\left( {{{{t^{{1 \over 7}}}} \over {{1 \over 7}}}} \right)$$ + C
= $${\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}}$$ + C
= $$\int {{{dx} \over {{{\left( {x + 4} \right)}^2}{{\left( {{{x - 3} \over {x + 4}}} \right)}^{{6 \over 7}}}}}} $$
Put $${{{x - 3} \over {x + 4}}}$$ = t
$$ \Rightarrow $$ $$\left\{ {{{\left( {x + 4} \right) - \left( {x - 3} \right)} \over {{{\left( {x + 4} \right)}^2}}}} \right\}dx$$ = dt
$$ \Rightarrow $$ $${{dx} \over {{{\left( {x + 4} \right)}^2}}} = {{dt} \over 7}$$
= $${1 \over 7}\int {{{dt} \over {{{\left( t \right)}^{{6 \over 7}}}}}} $$
= $${1 \over 7}\left( {{{{t^{{1 \over 7}}}} \over {{1 \over 7}}}} \right)$$ + C
= $${\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}}$$ + C
Comments (0)
