JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 19)
If $$\int {{{2{e^x} + 3{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}dx = {1 \over {14}}(ux + v{{\log }_e}(4{e^x} + 7{e^{ - x}})) + C} $$, where C is a constant of integration, then u + v is equal to _____________.
Answer
7
Explanation
$$2{e^x} + 3{e^{ - x}} = A(4{e^x} + 7{e^{ - x}}) + B(4{e^x} - 7{e^{ - x}}) + \lambda $$
2 = 4A + 4B ; 3 = 7A $$-$$ 7B ; $$\lambda$$ = 0
$$A + B = {1 \over 2}$$
$$A - B = {3 \over 7}$$
$$A = {1 \over 2}\left( {{1 \over 2} + {3 \over 7}} \right) = {{7 + 6} \over {28}} = {{13} \over {28}}$$
$$B = A - {3 \over 7} = {{13} \over {28}} - {3 \over 7} = {{13 - 12} \over {28}} = {1 \over {28}}$$
$$\int {{{13} \over {28}}dx + {1 \over {28}}\int {{{4{e^x} - 7{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}} dx} $$
= $${{13} \over {28}}x + {1 \over {28}}\ln |4{e^x} + 7{e^{ - x}}| + C$$
$$u = {{13} \over 2};v = {1 \over 2}$$
$$\Rightarrow$$ u + v = 7
2 = 4A + 4B ; 3 = 7A $$-$$ 7B ; $$\lambda$$ = 0
$$A + B = {1 \over 2}$$
$$A - B = {3 \over 7}$$
$$A = {1 \over 2}\left( {{1 \over 2} + {3 \over 7}} \right) = {{7 + 6} \over {28}} = {{13} \over {28}}$$
$$B = A - {3 \over 7} = {{13} \over {28}} - {3 \over 7} = {{13 - 12} \over {28}} = {1 \over {28}}$$
$$\int {{{13} \over {28}}dx + {1 \over {28}}\int {{{4{e^x} - 7{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}} dx} $$
= $${{13} \over {28}}x + {1 \over {28}}\ln |4{e^x} + 7{e^{ - x}}| + C$$
$$u = {{13} \over 2};v = {1 \over 2}$$
$$\Rightarrow$$ u + v = 7
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