What is the integral of 4e^xsec(e^x+1) ?

The integral of 4e^xsec(e^x+1) is 4ln|tan(e^x+1)+sec(e^x+1)|+C

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integral of 4e^xsec(e^x+1)

\int 4 e^{x} sec (e^{x} + 1) d x

4 ln ∣ tan (e^{x} + 1) + sec (e^{x} + 1) ∣ + C

Solution steps

\int 4 e^{x} sec (e^{x} + 1) d x

Take the constant out:

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

= 4 \cdot \int e^{x} sec (e^{x} + 1) d x

Apply u-substitution

: \int sec (u) d u

= 4 \cdot \int sec (u) d u

Use the common integral:

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

= 4 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = e^{x} + 1

= 4 ln ∣ tan (e^{x} + 1) + sec (e^{x} + 1) ∣

Add a constant to the solution

= 4 ln ∣ tan (e^{x} + 1) + sec (e^{x} + 1) ∣ + C

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What is the integral of 4e^xsec(e^x+1) ?
The integral of 4e^xsec(e^x+1) is 4ln|tan(e^x+1)+sec(e^x+1)|+C