What is the integral of 5e^xsec(e^x+2) ?

The integral of 5e^xsec(e^x+2) is 5ln|tan(e^x+2)+sec(e^x+2)|+C

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integral of 5e^xsec(e^x+2)

\int 5 e^{x} sec (e^{x} + 2) d x

5 ln ∣ tan (e^{x} + 2) + sec (e^{x} + 2) ∣ + C

Solution steps

\int 5 e^{x} sec (e^{x} + 2) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int e^{x} sec (e^{x} + 2) d x

Apply u-substitution

: \int sec (u) d u

= 5 \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 5 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = e^{x} + 2

= 5 ln ∣ tan (e^{x} + 2) + sec (e^{x} + 2) ∣

Add a constant to the solution

= 5 ln ∣ tan (e^{x} + 2) + sec (e^{x} + 2) ∣ + C

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What is the integral of 5e^xsec(e^x+2) ?
The integral of 5e^xsec(e^x+2) is 5ln|tan(e^x+2)+sec(e^x+2)|+C