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.

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Set $2x+1=u\implies 2dx=du$

$$\int\sec(2x+1)\ dx=\frac12\int\sec u\ du=\frac\ln2+K$$

Replace back $u$ with $2x+1$




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$$\frac 12\ln|\sec(2x+1)+\tan(2x+1)|+\text C$$$$\colorgreen+\text C$$

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