What is the integral of (18)/(18+e^x) ?

The integral of (18)/(18+e^x) is x-ln|e^x+18|+C

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integral of (18)/(18+e^x)

\int \frac{18}{18 + e ^{x}} d x

x - ln ∣ e^{x} + 18 ∣ + C

Solution steps

\int \frac{18}{18 + e ^{x}} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{18 + e ^{x}} d x

Apply u-substitution

= 18 \cdot \int \frac{1}{u ( 18 + u )} d u

Take the partial fraction of

\frac{1}{u ( 18 + u )} : \frac{1}{18 u} - \frac{1}{18 ( u + 18 )}

= 18 \cdot \int \frac{1}{18 u} - \frac{1}{18 ( u + 18 )} 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

= 18 (\int \frac{1}{18 u} d u - \int \frac{1}{18 ( u + 18 )} d u)

\int \frac{1}{18 u} d u = \frac{1}{18} ln ∣ u ∣

\int \frac{1}{18 ( u + 18 )} d u = \frac{1}{18} ln ∣ u + 18 ∣

= 18 (\frac{1}{18} ln ∣ u ∣ - \frac{1}{18} ln ∣ u + 18 ∣)

Substitute back

u = e^{x}

= 18 (\frac{1}{18} ln ∣ e^{x} ∣ - \frac{1}{18} ln ∣ e^{x} + 18 ∣)

Simplify

18 (\frac{1}{18} ln ∣ e^{x} ∣ - \frac{1}{18} ln ∣ e^{x} + 18 ∣) : x - ln ∣ e^{x} + 18 ∣

= x - ln ∣ e^{x} + 18 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 18 ∣ + C

d er i v a t i v e 3^{x^{3}}

d er i v a t i v e p (x) = (x - 3)^{3} - (x - 2)^{2} + 31

\int_{\frac{1}{2}}^{1} \frac{x + 4}{x ^{2} + x} d x

\int_{0}^{1} \int 1 d x d y

x \to 4 lim (x + 2)

What is the integral of (18)/(18+e^x) ?
The integral of (18)/(18+e^x) is x-ln|e^x+18|+C