Integrate the function: $\frac{1}{\sqrt{1+4x^2}}$

Let $2x = t$.
Then,$2 dx = dt$,which implies $dx = \frac{1}{2} dt$.
Substituting these into the integral:
$\int \frac{1}{\sqrt{1+4x^2}} dx = \int \frac{1}{\sqrt{1+t^2}} \cdot \frac{1}{2} dt = \frac{1}{2} \int \frac{1}{\sqrt{1+t^2}} dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{x^2+a^2}} dx = \log |x + \sqrt{x^2+a^2}| + C$:
$= \frac{1}{2} \log |t + \sqrt{t^2+1}| + C$.
Substituting $t = 2x$ back into the expression:
$= \frac{1}{2} \log |2x + \sqrt{4x^2+1}| + C$,where $C$ is an arbitrary constant.