integration of e^2logsecx?

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I am assuming log is natural logarithm or ln instead of log₁₀

∫ e^( 2 ln(sec(x) ) ) dx

Remember laws of natural logarithms and laws of exponents;
Laws of Exponents: =========> e^(2a) = e^a * e^a
Laws of Natural Logarithm: ===> e^ln(b) = b

Laws of Exponents:
e^( 2 ln(sec(x) ) ) = e^( ln(sec(x) ) * e^( ln(sec(x) )

Laws of Natural Logarithm:
e^( ln(sec(x) ) * e^( ln(sec(x) ) = sec(x) * sec(x) = sec²(x)

In short ∫ e^( 2 ln(sec(x) ) ) dx = ∫ sec²(x) dx

∫ e^( 2 ln(sec(x) ) ) dx
= ∫ sec²(x) dx and this is a common integral
= tan(x) + C
━━━━━━

e^(2 log sec x) = [e^(log sec x)]^2 = (sec x)^2

The integral of sec^2 x is tan x + C.

Hint: Its right there in your derivatives tables that you learned a while back. Just try to remember!

Or, look at your integrale tables. You can also figure it out by substitution.

Done!

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.