What is the integral of (14)/(14+e^x) ?

The integral of (14)/(14+e^x) is x-ln|e^x+14|+C

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integral of (14)/(14+e^x)

\int \frac{14}{14 + e ^{x}} d x

x - ln ∣ e^{x} + 14 ∣ + C

Solution steps

\int \frac{14}{14 + e ^{x}} d x

Take the constant out:

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

= 14 \cdot \int \frac{1}{14 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 14 + u )} d u

= 14 \cdot \int \frac{1}{u ( 14 + u )} d u

Take the partial fraction of

\frac{1}{u ( 14 + u )} : \frac{1}{14 u} - \frac{1}{14 ( u + 14 )}

= 14 \cdot \int \frac{1}{14 u} - \frac{1}{14 ( u + 14 )} 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

= 14 (\int \frac{1}{14 u} d u - \int \frac{1}{14 ( u + 14 )} d u)

\int \frac{1}{14 u} d u = \frac{1}{14} ln ∣ u ∣

\int \frac{1}{14 ( u + 14 )} d u = \frac{1}{14} ln ∣ u + 14 ∣

= 14 (\frac{1}{14} ln ∣ u ∣ - \frac{1}{14} ln ∣ u + 14 ∣)

Substitute back

u = e^{x}

= 14 (\frac{1}{14} ln ∣ e^{x} ∣ - \frac{1}{14} ln ∣ e^{x} + 14 ∣)

Simplify

14 (\frac{1}{14} ln ∣ e^{x} ∣ - \frac{1}{14} ln ∣ e^{x} + 14 ∣) : x - ln ∣ e^{x} + 14 ∣

= x - ln ∣ e^{x} + 14 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 14 ∣ + C

f (t) = t^{3} - 2 t^{2} + 5 t + 4

f (x) = \frac{e ^{- x}}{x + 1}, at x = 1

\frac{d}{d x} (\frac{x}{( 1 - x )})

\int x^{3} x - 4 d x

\frac{d}{d x} (\frac{2 x ^{2} - 3 x + 1}{2 x + 1})

What is the integral of (14)/(14+e^x) ?
The integral of (14)/(14+e^x) is x-ln|e^x+14|+C