Integrate 1 / (1 + 2x + 4x2) dx

The denominator is $4x^2 + 2x + 1$.

$$4x^2 + 2x + 1 = 4\left(x^2 + \frac{1}{2}x + \frac{1}{4}\right)$$

To complete the square for $x^2 + \frac{1}{2}x$, we add and subtract $\left(\frac{1}{2} \cdot \frac{1}{2}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.

$$4\left(x^2 + \frac{1}{2}x + \frac{1}{16} - \frac{1}{16} + \frac{1}{4}\right)$$ $$4\left(\left(x + \frac{1}{4}\right)^2 - \frac{1}{16} + \frac{4}{16}\right)$$ $$4\left(\left(x + \frac{1}{4}\right)^2 + \frac{3}{16}\right)$$

Let $u = x + \frac{1}{4}$. Then $du = dx$.
The integral becomes:

$$\frac{1}{4} \int \frac{1}{u^2 + \frac{3}{16}} du$$

This is in the form $\int \frac{1}{u^2+a^2} du$, where $a^2 = \frac{3}{16}$, so $a = \sqrt{\frac{3}{16}} = \frac{\sqrt{3}}{4}$.

The formula is $\int \frac{1}{u^2+a^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$.

$$\frac{1}{4} \left[ \frac{1}{\frac{\sqrt{3}}{4}} \arctan\left(\frac{u}{\frac{\sqrt{3}}{4}}\right) \right] + C$$ $$\frac{1}{4} \left[ \frac{4}{\sqrt{3}} \arctan\left(\frac{4u}{\sqrt{3}}\right) \right] + C$$ $$\frac{1}{\sqrt{3}} \arctan\left(\frac{4u}{\sqrt{3}}\right) + C$$

$$\frac{1}{\sqrt{3}} \arctan\left(\frac{4\left(x + \frac{1}{4}\right)}{\sqrt{3}}\right) + C$$ $$\frac{1}{\sqrt{3}} \arctan\left(\frac{4x + 1}{\sqrt{3}}\right) + C$$

The final answer is $\boxed{\frac{1}{\sqrt{3}} \arctan\left(\frac{4x + 1}{\sqrt{3}}\right) + C}$.