What is the integral of 1/(e^x+2) ?

The integral of 1/(e^x+2) is 1/2 x-1/2 ln|e^x+2|+C

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integral of 1/(e^x+2)

\int \frac{1}{e ^{x} + 2} d x

\frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣ + C

Solution steps

\int \frac{1}{e ^{x} + 2} d x

Apply u-substitution

: \int \frac{1}{u ( u + 2 )} d u

= \int \frac{1}{u ( u + 2 )} d u

Take the partial fraction of

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

= \int \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2 u} d u - \int \frac{1}{2 ( u + 2 )} d u

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣

Substitute back

u = e^{x}

= \frac{1}{2} ln ∣ e^{x} ∣ - \frac{1}{2} ln ∣ e^{x} + 2 ∣

\frac{1}{2} ln ∣ e^{x} ∣ = \frac{1}{2} x

= \frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣

Add a constant to the solution

= \frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣ + C

\frac{d}{d x} (e^{- x^{2}} - sin (x) cos (x))

\frac{\partial}{\partial x} (ln (2 x^{3} + y))

f (x) = (- 25 x^{2} + 9)^{2}

\frac{\partial}{\partial x} (1 - x - y)

\frac{4}{s + 2}

What is the integral of 1/(e^x+2) ?
The integral of 1/(e^x+2) is 1/2 x-1/2 ln|e^x+2|+C