What is the integral of (15)/(15+e^x) ?

The integral of (15)/(15+e^x) is x-ln|e^x+15|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (15)/(15+e^x)

\int \frac{15}{15 + e ^{x}} d x

x - ln ∣ e^{x} + 15 ∣ + C

Solution steps

\int \frac{15}{15 + e ^{x}} d x

Take the constant out:

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

= 15 \cdot \int \frac{1}{15 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 15 + u )} d u

= 15 \cdot \int \frac{1}{u ( 15 + u )} d u

Take the partial fraction of

\frac{1}{u ( 15 + u )} : \frac{1}{15 u} - \frac{1}{15 ( u + 15 )}

= 15 \cdot \int \frac{1}{15 u} - \frac{1}{15 ( u + 15 )} 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

= 15 (\int \frac{1}{15 u} d u - \int \frac{1}{15 ( u + 15 )} d u)

\int \frac{1}{15 u} d u = \frac{1}{15} ln ∣ u ∣

\int \frac{1}{15 ( u + 15 )} d u = \frac{1}{15} ln ∣ u + 15 ∣

= 15 (\frac{1}{15} ln ∣ u ∣ - \frac{1}{15} ln ∣ u + 15 ∣)

Substitute back

u = e^{x}

= 15 (\frac{1}{15} ln ∣ e^{x} ∣ - \frac{1}{15} ln ∣ e^{x} + 15 ∣)

Simplify

15 (\frac{1}{15} ln ∣ e^{x} ∣ - \frac{1}{15} ln ∣ e^{x} + 15 ∣) : x - ln ∣ e^{x} + 15 ∣

= x - ln ∣ e^{x} + 15 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 15 ∣ + C

\int_{0}^{100} \frac{1}{x} d x

\int_{0}^{1} 2 x d x

y = 4^{x + 1}

6 \cdot y^{^{''}} + 2 \cdot y^{^{'}} = 0, y^{^{'}} (0) = 4

\int_{0}^{1} \frac{2 ( x + 2 )}{5} d x

What is the integral of (15)/(15+e^x) ?
The integral of (15)/(15+e^x) is x-ln|e^x+15|+C