What is the integral of (36)/(36+e^x) ?

The integral of (36)/(36+e^x) is x-ln|e^x+36|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (36)/(36+e^x)

\int \frac{36}{36 + e ^{x}} d x

x - ln ∣ e^{x} + 36 ∣ + C

Solution steps

\int \frac{36}{36 + e ^{x}} d x

Take the constant out:

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

= 36 \cdot \int \frac{1}{36 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 36 + u )} d u

= 36 \cdot \int \frac{1}{u ( 36 + u )} d u

Take the partial fraction of

\frac{1}{u ( 36 + u )} : \frac{1}{36 u} - \frac{1}{36 ( u + 36 )}

= 36 \cdot \int \frac{1}{36 u} - \frac{1}{36 ( u + 36 )} 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

= 36 (\int \frac{1}{36 u} d u - \int \frac{1}{36 ( u + 36 )} d u)

\int \frac{1}{36 u} d u = \frac{1}{36} ln ∣ u ∣

\int \frac{1}{36 ( u + 36 )} d u = \frac{1}{36} ln ∣ u + 36 ∣

= 36 (\frac{1}{36} ln ∣ u ∣ - \frac{1}{36} ln ∣ u + 36 ∣)

Substitute back

u = e^{x}

= 36 (\frac{1}{36} ln ∣ e^{x} ∣ - \frac{1}{36} ln ∣ e^{x} + 36 ∣)

Simplify

36 (\frac{1}{36} ln ∣ e^{x} ∣ - \frac{1}{36} ln ∣ e^{x} + 36 ∣) : x - ln ∣ e^{x} + 36 ∣

= x - ln ∣ e^{x} + 36 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 36 ∣ + C

x \to 4 lim (\frac{3 x}{x - 4})

\frac{\partial}{\partial x} (e^{x y^{2}})

\frac{d}{d x} (\frac{6}{x + 3})

\frac{\frac{10}{6}}{s + 6}

\int_{0}^{2 π} cos (θ) d θ

What is the integral of (36)/(36+e^x) ?
The integral of (36)/(36+e^x) is x-ln|e^x+36|+C