Evaluate the integral of sec(2x + 1) dx I got $\ln|\sec(2x +1) + \tan(2x+1)| + \text C$ as an answer. I saw that the integral of $\sec x$ is $\ln|\sec x + \tan x| + \text C$. But I feel I may have left something out because that was too easy.

Let $u=2x+1$. Then $\text du=2$
$$\int \sec(2x+1) \ \text dx=\int \frac{\sec(u)}{2} \ \text du$$
$$=\frac 12\int \sec(u) \, \text du$$
$$=\frac 12\ln|\sec(u)+\tan(u)|+\text C$$
Reverse the substitution
$$\frac 12\ln|\sec(2x+1)+\tan(2x+1)|+\text C$$
$$\color{green}{\int \sec(2x+1) \, \text dx =\frac 12\ln|\sec(2x+1)+\tan(2x+1)|+\text C}$$
