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

The integral of 5/(e^x+2) is 5(1/2 x-1/2 ln|e^x+2|)+C

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integral of 5/(e^x+2)

\int \frac{5}{e ^{x} + 2} d x

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

Solution steps

\int \frac{5}{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 \frac{1}{e ^{x} + 2} d x

Apply u-substitution

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

= 5 \cdot \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 )}

= 5 \cdot \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

= 5 (\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 ∣

= 5 (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣)

Substitute back

u = e^{x}

= 5 (\frac{1}{2} ln ∣ e^{x} ∣ - \frac{1}{2} ln ∣ e^{x} + 2 ∣)

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

= 5 (\frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣)

Add a constant to the solution

= 5 (\frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣) + C

\int_{0}^{1} \frac{1}{2 - 3 x} d x

\frac{d y}{d x} + 3 x y^{3} = 0

\int \frac{5 t - 4}{t + 1} d t

x \to - 1 lim (0)

\int (1 - x)^{- 3} d x

What is the integral of 5/(e^x+2) ?
The integral of 5/(e^x+2) is 5(1/2 x-1/2 ln|e^x+2|)+C