$\int \frac{e^{x}}{2 e^{2 x} + 3 e^{x} + 2} d x is equal to$

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

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

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

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

answer is C.

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Detailed Solution

$\int \frac{e^{x}}{2 e^{2 x} + 3 e^{x} + 2} d x$

$Put e^{x} = t$

$\begin{array}{l} e^{x} d x - d t \\ \int \frac{e^{x}}{{(e^{x})}^{2} + 3 e^{x} + 2} d x = \int \frac{d t}{t^{2} + 3 t + 2} = \int \frac{d t}{t^{2} + 2 \frac{3}{2} t + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2} \end{array}$

$Use \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \log |\frac{x - a}{x + a}| + c$

$\begin{array}{l} = \int \frac{d t}{{(t + \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}} \\ = \frac{1}{2 \cdot \frac{1}{2}} \log |\frac{t + \frac{3}{2} - \frac{1}{2}}{t + \frac{3}{2} + \frac{1}{2}}| + k \\ = \log |\frac{2 e^{x} + 2}{2 e^{x} + 4}| + k = \log |\frac{e^{x} + 1}{e^{x} + 2}| + k \end{array}$