What is the integral of 4e^xsec(e^x+2) ?

The integral of 4e^xsec(e^x+2) is 4ln|tan(e^x+2)+sec(e^x+2)|+C

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integral of 4e^xsec(e^x+2)

\int 4 e^{x} sec (e^{x} + 2) d x

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

Solution steps

\int 4 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

= 4 \cdot \int e^{x} sec (e^{x} + 2) 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} + 2

= 4 ln ∣ tan (e^{x} + 2) + sec (e^{x} + 2) ∣

Add a constant to the solution

= 4 ln ∣ tan (e^{x} + 2) + sec (e^{x} + 2) ∣ + C

\int \frac{( x - 1 )}{( x - 1 ) ^{2} + 1} d x

\frac{1}{s ^{\frac{3}{2}}}

6 y^{^{''}} + 31 y = 0, y (0) = 6, y^{^{'}} (0) = 2

\frac{d}{d x} (\frac{1}{y ^{2} + x})

y = \frac{1}{x ^{2}}, (3, \frac{1}{9})

What is the integral of 4e^xsec(e^x+2) ?
The integral of 4e^xsec(e^x+2) is 4ln|tan(e^x+2)+sec(e^x+2)|+C