What is the integral from 0 to 5 of (e^x)/(5+e^x) ?

The integral from 0 to 5 of (e^x)/(5+e^x) is ln(5+e^5)-ln(6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 5 of (e^x)/(5+e^x)

\int_{0}^{5} \frac{e ^{x}}{5 + e ^{x}} d x

ln (5 + e^{5}) - ln (6)

Decimal Notation

3.24137 \dots

Solution steps

\int_{0}^{5} \frac{e ^{x}}{5 + e ^{x}} d x

Apply u-substitution

: \int_{6}^{5 + e^{5}} \frac{1}{u} d u

= \int_{6}^{5 + e^{5}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{6}^{5 + e^{5}}

Compute the boundaries:

ln (5 + e^{5}) - ln (6)

= ln (5 + e^{5}) - ln (6)

\int_{1}^{2} (x + 2 y)^{- 2} d x

\int_{3}^{5} x^{2} - 4 x d x

\int_{0.8}^{1} 5 (1 - x) d x

\int_{0}^{\frac{π}{3}} x sin (x + y) d y

\int_{x}^{2} \frac{x y}{2} d y

What is the integral from 0 to 5 of (e^x)/(5+e^x) ?
The integral from 0 to 5 of (e^x)/(5+e^x) is ln(5+e^5)-ln(6)