QJEEM20Ma6abf: JEE Main 2020 Mathematics - Class 12 JEEM20Ma6abf Solution: The integral is equal to : (where C is a constant of integration)

Question

Verified Solution • 2025–26 Session moderate

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)

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

Accepted Answer

\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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