∫9ex−4e−x4ex+6e−xdx=Ax+Blog(9e2x−4)+C,then A=_________ _, B=_________, C=___________

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{{{4}{e}^{{x}}+{6}{e}^{{-{x}}}}}{{{9}{e}^{{x}}-{4}{e}^{{-{x}}}

Question

$\int \frac{4 e ^{x} + 6 e ^{- x}}{9 e ^{x} - 4 e ^{- x}} d x = A x + B lo g (9 e^{2 x} - 4) + C, t h e n$ A=_________ _, B=_, C=___

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\int \frac{4 e ^{x} + 6 e ^{- x}}{9 e ^{x} - 4 e ^{- x}} d x = A x + B lo g (9 e^{2 x} - 4) + C, t h e n

A=_________ _, B=_________, C=___________

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Question Text

\int \frac{4 e ^{x} + 6 e ^{- x}}{9 e ^{x} - 4 e ^{- x}} d x = A x + B lo g (9 e^{2 x} - 4) + C, t h e n

A=_________ _, B=_________, C=___________