What is the integral of (19)/(19+e^x) ?

The integral of (19)/(19+e^x) is x-ln|e^x+19|+C

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integral of (19)/(19+e^x)

\int \frac{19}{19 + e ^{x}} d x

x - ln ∣ e^{x} + 19 ∣ + C

Solution steps

\int \frac{19}{19 + e ^{x}} d x

Take the constant out:

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

= 19 \cdot \int \frac{1}{19 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 19 + u )} d u

= 19 \cdot \int \frac{1}{u ( 19 + u )} d u

Take the partial fraction of

\frac{1}{u ( 19 + u )} : \frac{1}{19 u} - \frac{1}{19 ( u + 19 )}

= 19 \cdot \int \frac{1}{19 u} - \frac{1}{19 ( u + 19 )} 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

= 19 (\int \frac{1}{19 u} d u - \int \frac{1}{19 ( u + 19 )} d u)

\int \frac{1}{19 u} d u = \frac{1}{19} ln ∣ u ∣

\int \frac{1}{19 ( u + 19 )} d u = \frac{1}{19} ln ∣ u + 19 ∣

= 19 (\frac{1}{19} ln ∣ u ∣ - \frac{1}{19} ln ∣ u + 19 ∣)

Substitute back

u = e^{x}

= 19 (\frac{1}{19} ln ∣ e^{x} ∣ - \frac{1}{19} ln ∣ e^{x} + 19 ∣)

Simplify

19 (\frac{1}{19} ln ∣ e^{x} ∣ - \frac{1}{19} ln ∣ e^{x} + 19 ∣) : x - ln ∣ e^{x} + 19 ∣

= x - ln ∣ e^{x} + 19 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 19 ∣ + C

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x \to 0 lim (\frac{9 x ^{2} - 6 x}{x})

What is the integral of (19)/(19+e^x) ?
The integral of (19)/(19+e^x) is x-ln|e^x+19|+C