Evaluate the integral of tan2(x)sec(x).

Let $I = \int \sec^3 x \, dx$.
We use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \sec x$ and $dv = \sec^2 x \, dx$.
Then $du = \sec x \tan x \, dx$ and $v = \tan x$.

$$I = \sec x \tan x - \int \tan x (\sec x \tan x) \, dx$$ $$I = \sec x \tan x - \int \sec x \tan^2 x \, dx$$

Substitute $\tan^2 x = \sec^2 x - 1$ again:

$$I = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx$$ $$I = \sec x \tan x - \int (\sec^3 x - \sec x) \, dx$$ $$I = \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx$$ $$I = \sec x \tan x - I + \ln|\sec x + \tan x|$$

Now, solve for $I$:

$$2I = \sec x \tan x + \ln|\sec x + \tan x|$$ $$I = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x|$$

Substitute the expressions for $\int \sec^3 x \, dx$ and $\int \sec x \, dx$ back into the equation from Step 2:

$$\int \sec^3 x \, dx - \int \sec x \, dx = \left( \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| \right) - \left( \ln|\sec x + \tan x| \right) + C$$

$$\frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| - \ln|\sec x + \tan x| + C$$ $$\frac{1}{2} \sec x \tan x - \frac{1}{2} \ln|\sec x + \tan x| + C$$

The final answer is $\boxed{\frac{1}{2} \sec x \tan x - \frac{1}{2} \ln|\sec x + \tan x| + C}$.