If ∫2ex+3e−x3ex+4e−xdx=Ax+Blog⁡3e2x+4+C, then

A
$A = - \frac{3}{4}, B = \frac{1}{24}$
B
$A = \frac{3}{4}, B = - \frac{1}{24}$
C
$A = \frac{1}{4}, B = \frac{1}{24}$
D
$A = - \frac{3}{4}, B = \frac{1}{4}$

Solution:

Let

$\begin{aligned} \int \frac{2 e^{x} + 3 e^{- x}}{3 e^{x} + 4 e^{- x}} dx = \int \frac{2 e^{2 x} + 3}{3 e^{2 x} + 4} dx \\ 2 e^{2 x} + 3 = A (3 e^{2 x} + 4) + B (6 e^{2 x}) \\ \Rightarrow 2 e^{2 x} + 3 = (3 A + 6 B) e^{- 2 x} + 4 A \end{aligned}$

On comparing both sides, we get

$\begin{aligned} 2 = 3 A + 6 B - - - - (i) \\ 3 = 4 A - - - - (ii) \end{aligned}$

From Eqs. (i) and (ii), woe get

$\begin{array}{l} 2 = \frac{9}{4} + 6 B \\ \Rightarrow 6 B = 2 - \frac{9}{4} = - \frac{1}{4} \\ \Rightarrow B = - \frac{1}{24} \\ \begin{aligned} \Rightarrow \frac{2 e^{2 x} + 3}{3 e^{2 x} + 4} dx & = \int \frac{3 (3 e^{2 x} + 4)}{4 (3 e^{2 x} + 4)} dx - \frac{1}{24} \int \frac{6 e^{2 x}}{3 e^{2 x} + 4} dx \\ = \frac{3}{4} x - \frac{1}{24} \log (3 e^{2 x} + 4) + C \end{aligned} \end{array}$

If $\int \frac{2 e^{x} + 3 e^{- x}}{3 e^{x} + 4 e^{- x}} dx = Ax + Blog (3 e^{2 x} + 4) + C,$ then

Solution:

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