Evaluate the following integrals:

Let $\tan x=t$

Then $\mathrm{dt}=\sec ^{2} \mathrm{x} \mathrm{dx}$

Therefore, $\int \frac{\sec ^{2} x}{\sqrt{4+\tan ^{2} x}} d x=\int \frac{d t}{\sqrt{2^{2}+t^{2}}}$

Since, $\int \frac{1}{\sqrt{\left(x^{2}+a^{2}\right)}} d x=\log \left[x+\sqrt{\left(x^{2}\right.}+a^{2}\right)+c$

Hence, $\int \frac{d t}{\sqrt{2}^{2}+t^{2}}=\log \left[t+\sqrt{t^{2}+2^{2}}\right]+c$

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