Q57 / 90 · JEE Main 2021 (27 Aug Shift 2) · Mathematics

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

Correct answer: 7

Solution

I = \int \frac{2 e^{x} + 3 e^{- x}}{4 e^{x} + 7 e^{- x}} d x = \int \frac{2 e^{x} + 3}{4 e^{2 x} + 7} d x

Here

2 e^{2 x} + 3 = A (8 e^{2 x}) + B (4 e^{2 x} + 7)

\Rightarrow 2 e^{2 x} + 3 = (8 A + 4 B) e^{2 x} + 7 B

\Rightarrow B = \frac{3}{7} & A = \frac{1}{28}

I = \int \frac{\frac{1}{28} (8 e^{2 x}) + \frac{3}{7} (4 e^{2 x} + 7)}{4 e^{2 x} + 7} d x

I = \frac{1}{28} \ln |4 e^{2 x} + 7| + \frac{3}{7} x + C

I = \frac{1}{28} \ln ∣ e^{x} (4 e^{x} + 7 e^{- x} ∣ + \frac{3}{7} x + C

\Rightarrow u = \frac{13}{2} & v = \frac{1}{2}

\Rightarrow u + v = \frac{13}{2} + \frac{1}{2} = 7

Hence, the correct answer is 7.

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