Let P(x) = ∫ dx e^ x + 8e^ - x + 4e^ - 3x , Q(x) = ∫ dx e^ 3x + 8e^...

We have, $R\left(x\right) = ∫ \dfrac{dx}{e^{x} + 8e^{- x} + 4e^{- 3x}} - 2 ∫ \dfrac{dx}{e^{3x} + 8e^{x} + 4e^{- x}}$

$⇒ R\left(x\right) = ∫ \dfrac{e^{x}\left(e^{2x} - 2\right)}{e^{4x} + 8e^{2x} + 4}dx$

we get $R\left(t\right) = ∫ \dfrac{\left(t^{2} - 2\right)dt}{t^{4} + 8t^{2} + 4} = ∫ \dfrac{\left(1 - 2t^{- 2}\right)dt}{\left(t + 2t^{- 1}\right)^{2} + 4}$

$= \dfrac{1}{2}\tan^{-1}\left(\dfrac{t + 2t^{- 1}}{2}\right) + K$

$⇒ R\left(x\right) = \dfrac{1}{2}\tan^{-1}\left(\dfrac{e^{x} + 2e^{- x}}{2}\right) + K$

Hence, \left(A, B, C\right) =\left(\tan^{-1},e^{x}, 2\right)$</p><p>$∴ Option (b) is correct.