What is the integral of 3/(3+e^x) ?

The integral of 3/(3+e^x) is x-ln|e^x+3|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3/(3+e^x)

\int \frac{3}{3 + e ^{x}} d x

x - ln ∣ e^{x} + 3 ∣ + C

Solution steps

\int \frac{3}{3 + e ^{x}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{3 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 3 + u )} d u

= 3 \cdot \int \frac{1}{u ( 3 + u )} d u

Take the partial fraction of

\frac{1}{u ( 3 + u )} : \frac{1}{3 u} - \frac{1}{3 ( u + 3 )}

= 3 \cdot \int \frac{1}{3 u} - \frac{1}{3 ( u + 3 )} 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

= 3 (\int \frac{1}{3 u} d u - \int \frac{1}{3 ( u + 3 )} d u)

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u + 3 )} d u = \frac{1}{3} ln ∣ u + 3 ∣

= 3 (\frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 3 ∣)

Substitute back

u = e^{x}

= 3 (\frac{1}{3} ln ∣ e^{x} ∣ - \frac{1}{3} ln ∣ e^{x} + 3 ∣)

Simplify

3 (\frac{1}{3} ln ∣ e^{x} ∣ - \frac{1}{3} ln ∣ e^{x} + 3 ∣) : x - ln ∣ e^{x} + 3 ∣

= x - ln ∣ e^{x} + 3 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 3 ∣ + C

y^{^{''}} + y^{^{'}} - 30 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

(2 x + 1)^{^{'}}

y^{^{'}} = (y + 2 x)^{2}, y (0) = 3

4 x^{4} + 3 x^{3} + 2 x^{2} + x + 11

\frac{d}{d x} (\frac{- 1}{x})