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

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Solution

$2 e^{x} + 3 e^{- x} = A (4 e^{x} + 7 e^{- x}) + B (4 e^{x} - 7 e^{- x})$
Comparing both sides
$4 A + 4 B = 2 \dots (i)$
$7 A - 7 B = 3 \dots (i i)$
On solving, we get
$A = \frac{13}{28}$ and $B = \frac{1}{28}$
$I = \int \frac{2 e^{x} + 3 e^{- x}}{4 e^{x} + 7 e^{- x}} d x = \int ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{13}{28} (4 e^{x} + 7 e^{- x}) + \frac{1}{28} (4 e^{x} - 7 e^{- x})}{4 e^{x} + 7 e^{- x}} ⎞ ⎟ ⎟ ⎟ ⎠ d x$
$= \frac{13}{28} x + \frac{1}{28} {log}_{e} (4 e^{x} + 7 e^{- x}) + C$
Comparing LHS and RHS gives $u = \frac{13}{2}$ and $v = \frac{1}{2} \Rightarrow u + v = 7$

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