Expand the expression inside the square root.

$$
1 + 2\tan x (\tan x + \sec x) = 1 + 2\tan^2 x + 2\tan x \sec x
$$

$$
1 + 2\tan^2 x + 2\tan x \sec x = (1 + \tan^2 x) + \tan^2 x + 2\tan x \sec x
$$
$$
= \sec^2 x + \tan^2 x + 2\tan x \sec x
$$

Here, $a = \sec x$ and $b = \tan x$.
$$
\sec^2 x + \tan^2 x + 2\tan x \sec x = (\sec x + \tan x)^2
$$

$$
\int \sqrt{(\sec x + \tan x)^2}\,dx = \int |\sec x + \tan x|\,dx
$$
Assuming $\sec x + \tan x \ge 0$ for the domain of integration, we can remove the absolute value.
$$
\int (\sec x + \tan x)\,dx
$$

Recall that $\int \sec x\,dx = \ln|\sec x + \tan x| + C$ and $\int \tan x\,dx = \ln|\sec x| + C$.
$$
\int (\sec x + \tan x)\,dx = \ln|\sec x + \tan x| + \ln|\sec x| + C
$$

$$
\ln|\sec x (\sec x + \tan x)| + C = \ln|\sec^2 x + \sec x \tan x| + C
$$
$$
\boxed{\ln|\sec^2 x + \sec x \tan x| + C}
$$

$$
du = \frac{d}{dx}(x^2 - 2x + 26)\,dx = (2x - 2)\,dx = 2(x-1)\,dx
$$
From this, we have $(x-1)\,dx = \frac{1}{2}\,du$.

$$
\int \frac{1}{u} \cdot \frac{1}{2}\,du = \frac{1}{2} \int \frac{1}{u}\,du
$$

$$
\frac{1}{2} \ln|u| + C
$$

The discriminant of $x^2 - 2x + 26$ is $(-2)^2 - 4(1)(26) = 4 - 104 = -100$, which is negative. Since the leading coefficient is positive, $x^2 - 2x + 26$ is always positive, so $|x^2 - 2x + 26| = x^2 - 2x + 26$.
$$
\boxed{\frac{1}{2} \ln(x^2 - 2x + 26) + C}
$$

The general form is $\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\left(\frac{x}{a}\right) + C$.

In this integral, $a^2 = 4$, so $a = 2$.

$$
\int \frac{1}{\sqrt{2^2 - x^2}}\,dx = \arcsin\left(\frac{x}{2}\right) + C
$$
$$
\boxed{\arcsin\left(\frac{x}{2}\right) + C}
$$