What is the integral of (32)/(32+e^x) ?

The integral of (32)/(32+e^x) is x-ln|e^x+32|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (32)/(32+e^x)

\int \frac{32}{32 + e ^{x}} d x

x - ln ∣ e^{x} + 32 ∣ + C

Solution steps

\int \frac{32}{32 + e ^{x}} d x

Take the constant out:

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

= 32 \cdot \int \frac{1}{32 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 32 + u )} d u

= 32 \cdot \int \frac{1}{u ( 32 + u )} d u

Take the partial fraction of

\frac{1}{u ( 32 + u )} : \frac{1}{32 u} - \frac{1}{32 ( u + 32 )}

= 32 \cdot \int \frac{1}{32 u} - \frac{1}{32 ( u + 32 )} 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

= 32 (\int \frac{1}{32 u} d u - \int \frac{1}{32 ( u + 32 )} d u)

\int \frac{1}{32 u} d u = \frac{1}{32} ln ∣ u ∣

\int \frac{1}{32 ( u + 32 )} d u = \frac{1}{32} ln ∣ u + 32 ∣

= 32 (\frac{1}{32} ln ∣ u ∣ - \frac{1}{32} ln ∣ u + 32 ∣)

Substitute back

u = e^{x}

= 32 (\frac{1}{32} ln ∣ e^{x} ∣ - \frac{1}{32} ln ∣ e^{x} + 32 ∣)

Simplify

32 (\frac{1}{32} ln ∣ e^{x} ∣ - \frac{1}{32} ln ∣ e^{x} + 32 ∣) : x - ln ∣ e^{x} + 32 ∣

= x - ln ∣ e^{x} + 32 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 32 ∣ + C

x^{\frac{3}{2}}

x \to 7 lim (2 - x - 3)

\int e^{x} 39 + e^{x} d x

f (x) = x^{3} + 3, (1, 2)

\int - 3 x d x

What is the integral of (32)/(32+e^x) ?
The integral of (32)/(32+e^x) is x-ln|e^x+32|+C