What is the integral of (17)/(17+e^x) ?

The integral of (17)/(17+e^x) is x-ln|e^x+17|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (17)/(17+e^x)

\int \frac{17}{17 + e ^{x}} d x

x - ln ∣ e^{x} + 17 ∣ + C

Solution steps

\int \frac{17}{17 + e ^{x}} d x

Take the constant out:

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

= 17 \cdot \int \frac{1}{17 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 17 + u )} d u

= 17 \cdot \int \frac{1}{u ( 17 + u )} d u

Take the partial fraction of

\frac{1}{u ( 17 + u )} : \frac{1}{17 u} - \frac{1}{17 ( u + 17 )}

= 17 \cdot \int \frac{1}{17 u} - \frac{1}{17 ( u + 17 )} 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

= 17 (\int \frac{1}{17 u} d u - \int \frac{1}{17 ( u + 17 )} d u)

\int \frac{1}{17 u} d u = \frac{1}{17} ln ∣ u ∣

\int \frac{1}{17 ( u + 17 )} d u = \frac{1}{17} ln ∣ u + 17 ∣

= 17 (\frac{1}{17} ln ∣ u ∣ - \frac{1}{17} ln ∣ u + 17 ∣)

Substitute back

u = e^{x}

= 17 (\frac{1}{17} ln ∣ e^{x} ∣ - \frac{1}{17} ln ∣ e^{x} + 17 ∣)

Simplify

17 (\frac{1}{17} ln ∣ e^{x} ∣ - \frac{1}{17} ln ∣ e^{x} + 17 ∣) : x - ln ∣ e^{x} + 17 ∣

= x - ln ∣ e^{x} + 17 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 17 ∣ + C

\int_{0}^{\infty} x e^{- 5 x} d x

\frac{d}{d x} (2^{t a n (4 x)} + \frac{csc ^{3} ( x )}{ln ( x ^{2} )})

x \to - 1 lim ((sin (\frac{π}{2}) x + 2^{- 3 x})^{2})

\int 29 tan^{5} (x) sec^{4} (x) d x

\int_{0}^{4} 7 e^{- x^{2}} d x

What is the integral of (17)/(17+e^x) ?
The integral of (17)/(17+e^x) is x-ln|e^x+17|+C