JEE MAIN - Mathematics (2015 (Offline) - No. 9)
The integral $$\int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} $$ equals :
$$ - {\left( {{x^4} + 1} \right)^{{1 \over 4}}} + c$$
$$ - {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{{1 \over 4}}} + c$$
$$ {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{{1 \over 4}}} + c$$
$$ {\left( {{x^4} + 1} \right)^{{1 \over 4}}} + c$$
Explanation
$$1 = \int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} $$
$$ = \int {{{dx} \over {{x^3}{{\left( {1 + {x^{ - 4}}} \right)}^{3/4}}}}} $$
Let $${x^{ - 4}} = y$$
$$ \Rightarrow - 4{x^{ - 3}}\,dx = dy$$
$$ \Rightarrow dx = {{ - 1} \over 4}{x^3}dy$$
$$\therefore$$ $$I = {{ - 1} \over 4}\int {{{{x^3}dy} \over {{x^3}{{\left( {1 + y} \right)}^{3/4}}}}} $$
$$ = {{ - 1} \over 4}\int {{{dy} \over {{{\left( {1 + y} \right)}^{3/4}}}}} $$
$$ = {{ - 1} \over 4} \times 4{\left( {1 + y} \right)^{1/4}}$$
$$ = - 1{\left( {1 + {x^{ - 4}}} \right)^{1/4}} + C$$
$$ = - {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{1/4}} + C$$
$$ = \int {{{dx} \over {{x^3}{{\left( {1 + {x^{ - 4}}} \right)}^{3/4}}}}} $$
Let $${x^{ - 4}} = y$$
$$ \Rightarrow - 4{x^{ - 3}}\,dx = dy$$
$$ \Rightarrow dx = {{ - 1} \over 4}{x^3}dy$$
$$\therefore$$ $$I = {{ - 1} \over 4}\int {{{{x^3}dy} \over {{x^3}{{\left( {1 + y} \right)}^{3/4}}}}} $$
$$ = {{ - 1} \over 4}\int {{{dy} \over {{{\left( {1 + y} \right)}^{3/4}}}}} $$
$$ = {{ - 1} \over 4} \times 4{\left( {1 + y} \right)^{1/4}}$$
$$ = - 1{\left( {1 + {x^{ - 4}}} \right)^{1/4}} + C$$
$$ = - {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{1/4}} + C$$
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