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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)

JEE · Math · previous-year question

  1. A.$${1 \over 2}{\left( {{{x - 3} \over {x + 4}}} \right)^{{3 \over 7}}} + C$$
  2. B.$${\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}} + C$$correct
  3. C.$$ - {1 \over {13}}{\left( {{{x - 3} \over {x + 4}}} \right)^{{{13} \over 7}}} + C$$
  4. D.-$${\left( {{{x - 3} \over {x + 4}}} \right)^{-{1 \over 7}}} + C$$

Answer

B. $${\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}} + C$$

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

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