Evaluate the integral of sec(x) dx.

The integral of sec(x) dx is ln|sec(x) + tan(x)| + C.

To evaluate the integral of sec(x) dx, we can use the substitution method. Let u = sec(x) + tan(x), then du/dx = sec(x)tan(x) + sec^2(x) = u(tan(x) + u). Rearranging, we get dx = du/(tan(x) + u).

Substituting u and dx into the integral, we get:

∫sec(x) dx = ∫du/(tan(x) + u)

We can now use partial fractions to split the integral into two parts:

1/(tan(x) + u) = A/tan(x) + B/u

Multiplying both sides by (tan(x) + u) and setting u = -tan(x), we get:

A = -1/tan(x) = -cot(x)

Multiplying both sides by u and setting u = 1, we get:

B = 1

Therefore, we have:

1/(tan(x) + u) = -cot(x)/tan(x) + 1/u

Substituting this back into the integral, we get:

∫sec(x) dx = ∫(-cot(x)/tan(x) + 1/u) du

= -ln|tan(x)| + ln|sec(x) + tan(x)| + C

Simplifying, we get:

∫sec(x) dx = ln|sec(x) + tan(x)| - ln|cos(x)| + C

= ln|sec(x) + tan(x)| + ln|sec(x)| + C

= ln|sec(x)(sec(x) + tan(x))| + C

= ln|sec(x) + tan(x)| + C

Therefore, the integral of sec(x) dx is ln|sec(x) + tan(x)| + C.