text{If } int frac{2 e^{x}+3 e^{-x}}{4 e^{x}+ | vedclass

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(C) Let $I = \int \frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x$.We express the numerator as $2 e^{x}+3 e^{-x} = A(4 e^{x}+7 e^{-x}) + B(4 e^{x}-7 e^{

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Let $I = \int \frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x$.We express the numerator as $2 e^{x}+3 e^{-x} = A(4 e^{x}+7 e^{-x}) + B(4 e^{x}-7 e^{-x})$.Comparing coefficients of $e^{x}$ and $e^{-x}$:$4A + 4B = 2 \Rightarrow A+B = \frac{1}{2}$$7A - 7B = 3 \Rightarrow A-B = \frac{3}{7}$Adding the two equations: $2A = \frac{1}{2} + \frac{3}{7} = \frac{7+6}{14} = \frac{13}{14} \Rightarrow A = \frac{13}{28}$.Subtracting the two equations: $2B = \frac{1}{2} - \frac{3}{7} = \frac{7-6}{14} = \frac{1}{14} \Rightarrow B = \frac{1}{28}$.Thus,$I = \int \left( \frac{13}{28} + \frac{1}{28} \frac{4 e^{x}-7 e^{-x}}{4 e^{x}+7 e^{-x}} \right) d x$.$I = \frac{13}{28} x + \frac{1}{28} \log_{e} |4 e^{x}+7 e^{-x}| + C$.To match the form $\frac{1}{14}(u x + v \log_{e}(4 e^{x}+7 e^{-x})) + C$,we rewrite $I$ as:$I = \frac{1}{14} (\frac{13}{2} x + \frac{1}{2} \log_{e}(4 e^{x}+7 e^{-x})) + C$.Comparing,we get $u = \frac{13}{2}$ and $v = \frac{1}{2}$.Therefore,$u+v = \frac{13}{2} + \frac{1}{2} = \frac{14}{2} = 7$.

$\text{If } \int \frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x=\frac{1}{14}\left(u x+v \log _{e}\left(4 e^{x}+7 e^{-x}\right)\right)+C$ where $C$ is a constant of integration,then $u+v$ is equal to .... .

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