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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$\begingroup$Set $2x+1=u\implies 2dx=du$
$$\int\sec(2x+1)\ dx=\frac12\int\sec u\ du=\frac{\ln|\sec u+\tan u|}2+K$$
Replace back $u$ with $2x+1$
$\endgroup$ $\begingroup$You neglected the chain rule: $$ \frac{d}{dx} \ln|\sec(2x+1)+\tan(2x+1)| = \sec(2x+1)\cdot 2. $$
$\endgroup$ 1 $\begingroup$$$\int\sec({2x+1})\,dx=\frac{1}{2}\int\sec({2x+1})\,d(2x+1)$$ Can you take it from here?
$\endgroup$ $\begingroup$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 $$\frac 12\ln|\sec(2x+1)+\tan(2x+1)|+\text C$$ $$\color{green}{\int \sec(2x+1) \, \text dx =\frac 12\ln|\sec(2x+1)+\tan(2x+1)|+\text C}$$
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