Show step by step am slow learner

The denominator is $1+2x+4x^2$. It's usually easier to write the $x^2$ term first: $4x^2+2x+1$.
To complete the square for an expression like $Ax^2+Bx+C$, we first factor out $A$ from the terms involving $x$:

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

For an expression $x^2+bx$, to complete the square, we add and subtract $\left(\frac{b}{2}\right)^2$.
Here, $b = \frac{1}{2}$. So, $\frac{b}{2} = \frac{1/2}{2} = \frac{1}{4}$.
Then, $\left(\frac{b}{2}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
Now, we add and subtract $\frac{1}{16}$ inside the parenthesis:

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

Group the first three terms to form a perfect square:

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

Combine the constant terms: $-\frac{1}{16} + \frac{1}{4} = -\frac{1}{16} + \frac{4}{16} = \frac{3}{16}$.
So the denominator becomes:

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

Substitute the completed square back into the integral:

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

We can pull the constant $\frac{1}{4}$ out of the integral:

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

The integral form for arctangent is $\int \frac{1}{u^2+a^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$.
Let $u = x + \frac{1}{4}$. Then, the derivative $du = dx$.
From the denominator, we have $u^2 = \left(x + \frac{1}{4}\right)^2$.
And $a^2 = \frac{3}{16}$. To find $a$, we take the square root: $a = \sqrt{\frac{3}{16}} = \frac{\sqrt{3}}{\sqrt{16}} = \frac{\sqrt{3}}{4}$.
Now, substitute $u$ and $a$ into the integral:

$$\frac{1}{4} \int \frac{1}{u^2 + a^2} du = \frac{1}{4} \int \frac{1}{u^2 + \left(\frac{\sqrt{3}}{4}\right)^2} du$$

Using the formula $\int \frac{1}{u^2+a^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$:

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

Simplify the fraction $\frac{1}{\frac{\sqrt{3}}{4}}$:

$$\frac{1}{\frac{\sqrt{3}}{4}} = 1 \times \frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}}$$

So the expression becomes:

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

Multiply the terms:

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

Replace $u$ with $x + \frac{1}{4}$:

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

Distribute the 4 in the numerator of the arctangent argument:

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

So the final result is:

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