Answer
$$\int \frac{2 d x}{x \sqrt{1-4 \ln^{2} x}}=\sin^{-1}(2 \ln x)+C $$
Work Step by Step
Given $$\int \frac{2 d x}{x \sqrt{1-4 \ln^{2} x}}$$
Let $$\ln x=t \Rightarrow \frac{1}{x}dx= dt $$
So, we get
\begin{aligned} I&= \int \frac{2x d t}{x\sqrt{1-4 t^{2}}}\\
&=\int \frac{2 d t}{\sqrt{1-4 t^{2}}}\\
&=\int \frac{2 d t}{\sqrt{4 \left( \frac{1}{4}-t^{2}\right)}} \\
&=\int \frac{2 d t}{\sqrt{4}\sqrt{ \left( \frac{1}{4}-t^{2}\right)}} \\
&=\int \frac{ d t}{\sqrt{ \left( \frac{1}{4}-t^{2}\right)}} \\
&\text{Since}\int \frac{ d z}{\sqrt{ \left( a^2-z^{2}\right)}}=\sin^{-1}\frac{z}{a}, \text{So we get } \\
I&=\sin^{-1}\frac{t}{1/2}+C\\
&=\sin^{-1}2 t+C\\
&=\sin^{-1}(2 \ln x)+C
\end{aligned}
