Evaluate the following integrals:

Let $\mathrm{I}=\int \frac{1}{\sqrt{1+4 \mathrm{x}^{2}}} \mathrm{~d} \mathrm{x}=\int \frac{1}{\sqrt{1+(2 \mathrm{x})^{2}}} \mathrm{dx}$

Let $\mathrm{t}=2 \mathrm{x}$, then $\mathrm{dt}=2 \mathrm{dx}$ or $\mathrm{dx}=\mathrm{dt} / 2$

Therefore, $\int \frac{1}{\sqrt{1+(2 x)^{2}}} \mathrm{dx}=\frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{1+t^{2}}}$

$=\frac{1}{2} \log \left[\mathrm{t}+\sqrt{1+\mathrm{t}^{2}}\right]+\mathrm{c}\left\{\right.$ since $\left.\int \frac{1}{\sqrt{\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)}} \mathrm{dx}=\log \left[\mathrm{x}+\sqrt{\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)}+\mathrm{c}\right\}\right\}$

$=\frac{1}{2} \log \left[2 x+\sqrt{1+4 x^{2}}\right]+c$