$\int \frac{e^{x}}{(1 + e^{x}) (2 + e^{x})} dx$ is equal to

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$\log |\frac{2 + e^{x}}{1 + e^{x}}| + C$

$\log |\frac{1 + e^{x}}{1 - e^{x}}| + C$

$\frac{1}{2} \log |\frac{2 + e^{x}}{1 + e^{x}}| + C$

$\log |\frac{1 + e^{x}}{2 + e^{x}}| + C$

answer is D.

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Detailed Solution

Let, $I = \int \frac{e^{x}}{(1 + e^{x}) (2 + e^{x})} dx$

put, $e^{x} = t \Rightarrow e^{x} dx = dt$

$\begin{array}{l} ∴ I = \int \frac{e^{x}}{(1 + t) (2 + t)} \frac{dt}{e^{x}} = \int \frac{1}{(1 + t) (2 + t)} dt \\ Let \frac{1}{(1 + t) (2 + t)} = \frac{A}{(1 + t)} + \frac{B}{(2 + t)} \\ \Rightarrow 1 = A (2 + t) + B (1 + t) \\ = (2 A + B) + t (A + B) \end{array}$

On equating the coefficients of t and constant term on
both sides, we get $A + B = 0 and 2 A + B = 1$

$On solving both equations, we get A = 1 and B = - 1$

$\begin{array}{l} ∴ I = \int \frac{1}{(1 + t)} dt - \int \frac{1}{(2 + t)} dt \\ = \log | 1 + t | - \log | 2 + t | + C \\ = \log |\frac{1 + t}{2 + t}| + C = \log |\frac{1 + e^{x}}{2 + e^{x}}| + C \end{array}$