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QJEEM21Ma67d7: JEE Main 2021 Mathematics - Class 12 JEEM21Ma67d7 Solution: If , where C is a constant of integration, then u + v is equal…

If ∫2ex+3e−x4ex+7e−xdx=114(ux+vlog⁡e(4ex+7e−x))+C\int {{{2{e^x} + 3{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}dx = {1 \over {14}}(ux + v{{\log }_e}(4{e^x} + 7{e^{ - x}})) + C} , where C is a constant of integration, then u + v is equal to _____________.

Step-by-Step Verified Solution

2ex+3e−x=A(4ex+7e−x)+B(4ex−7e−x)+λ2{e^x} + 3{e^{ - x}} = A(4{e^x} + 7{e^{ - x}}) + B(4{e^x} - 7{e^{ - x}}) + \lambda

2 = 4A + 4B ; 3 = 7A −- 7B ; λ\lambda = 0

A+B=12A + B = {1 \over 2}

A−B=37A - B = {3 \over 7}

A=12(12+37)=7+628=1328A = {1 \over 2}\left( {{1 \over 2} + {3 \over 7}} \right) = {{7 + 6} \over {28}} = {{13} \over {28}}

B=A−37=1328−37=13−1228=128B = A - {3 \over 7} = {{13} \over {28}} - {3 \over 7} = {{13 - 12} \over {28}} = {1 \over {28}}

∫1328dx+128∫4ex−7e−x4ex+7e−xdx\int {{{13} \over {28}}dx + {1 \over {28}}\int {{{4{e^x} - 7{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}} dx}

= 1328x+128ln⁡∣4ex+7e−x∣+C{{13} \over {28}}x + {1 \over {28}}\ln |4{e^x} + 7{e^{ - x}}| + C

u=132;v=12u = {{13} \over 2};v = {1 \over 2}

⇒\Rightarrow u + v = 7

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