JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 7)
The integral $$\int {{1 \over {\root 4 \of {{{(x - 1)}^3}{{(x + 2)}^5}} }}} \,dx$$ is equal to : (where C is a constant of integration)
$${3 \over 4}{\left( {{{x + 2} \over {x - 1}}} \right)^{{1 \over 4}}} + C$$
$${3 \over 4}{\left( {{{x + 2} \over {x - 1}}} \right)^{{5 \over 4}}} + C$$
$${4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{{1 \over 4}}} + C$$
$${4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{{5 \over 4}}} + C$$
Explanation
$$\int {{{dx} \over {{{(x - 1)}^{3/4}}{{(x + 2)}^{5/4}}}}} $$
$$ = \int {{{dx} \over {{{\left( {{{x + 2} \over {x - 1}}} \right)}^{5/4}}.\,{{(x - 1)}^2}}}} $$
put $${{x + 2} \over {x - 1}} = t$$
$$ = - {1 \over 3}\int {{{dt} \over {{t^{5/4}}}}} $$
$$ = {4 \over 3}.{1 \over {{t^{1/4}}}} + C$$
$$ = {4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{1/4}} + C$$
$$ = \int {{{dx} \over {{{\left( {{{x + 2} \over {x - 1}}} \right)}^{5/4}}.\,{{(x - 1)}^2}}}} $$
put $${{x + 2} \over {x - 1}} = t$$
$$ = - {1 \over 3}\int {{{dt} \over {{t^{5/4}}}}} $$
$$ = {4 \over 3}.{1 \over {{t^{1/4}}}} + C$$
$$ = {4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{1/4}} + C$$
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