You are watching: Integral of 1/sec^2x
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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