$\int \frac{d x}{x [(\log x)^{2} + 4 \log x + 1}]$

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$\frac{1}{2 \sqrt{3}} \log |\frac{\log x + 2 - \sqrt{3}}{\log x + 2 + \sqrt{3}}| + c$

$\frac{1}{2 \sqrt{3}} \log |\frac{\log x + 2 + \sqrt{3}}{\log x + 2 - \sqrt{3}}| + c$

$\frac{1}{2 \sqrt{3}} \log |\frac{2 + \sqrt{3} + \log x}{2 + \sqrt{3} - \log x}| + c$

$\frac{1}{2 \sqrt{3}} \log |\frac{2 + \sqrt{3} - \log x}{2 + \sqrt{3} + \log x}| + c$

answer is A.

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Detailed Solution

$\int \frac{d x}{x [{(\log x)}^{2} + 4 \log x + 1]}$

$Put \log x = t$

$\begin{array}{l} \frac{1}{x} d x = d t \\ \int \frac{d x}{x [(\log x)^{2} + 4 \log x + 1]} = \int \frac{d t}{t^{2} + 4 t + 1} \end{array}$

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

$\begin{array}{l} = \int \frac{d t}{t^{2} + 2.2 t + (2)^{2} + 1 - (2)^{2}} \\ = \int \frac{d t}{(t + 2)^{2} - (3)} \\ = \int \frac{d t}{(t + 2)^{2} - (\sqrt{3})^{2}} \\ = \frac{1}{2 \sqrt{3}} \log |\frac{t + 2 - \sqrt{3}}{t + 2 + \sqrt{3}}| + c = \frac{1}{2 \sqrt{3}} \log |\frac{\log x + 2 - \sqrt{3}}{\log x + 2 + \sqrt{3}}| \end{array}$