Prove

Let $2 x=t$

$\therefore 2 d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{1+4 x^{2}}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{1+t^{2}}}$

$=\frac{1}{2}\left[\log \left|t+\sqrt{t^{2}+1}\right|\right]+\mathrm{C}$    $\left[\int \frac{1}{\sqrt{x^{2}+a^{2}}} d t=\log \left|x+\sqrt{x^{2}+a^{2}}\right|\right]$

$=\frac{1}{2} \log \left|2 x+\sqrt{4 x^{2}+1}\right|+\mathrm{C}$