What is the integral from 0 to 2 of (e^x)/(1+e^x) ?

The integral from 0 to 2 of (e^x)/(1+e^x) is ln(1+e^2)-ln(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of (e^x)/(1+e^x)

\int_{0}^{2} \frac{e ^{x}}{1 + e ^{x}} d x

ln (1 + e^{2}) - ln (2)

Decimal Notation

1.43378 \dots

Solution steps

\int_{0}^{2} \frac{e ^{x}}{1 + e ^{x}} d x

Apply u-substitution

: \int_{2}^{1 + e^{2}} \frac{1}{u} d u

= \int_{2}^{1 + e^{2}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{1 + e^{2}}

Compute the boundaries:

ln (1 + e^{2}) - ln (2)

= ln (1 + e^{2}) - ln (2)

\frac{d y}{d x} = \frac{x + 1}{y}

30 y^{^{''}} - 19 y^{^{'}} - 5 y = 0

x \to 7 lim (\frac{4}{( x - 7 ) ^{2}})

\int tan (x^{3}) x^{2} d x

\frac{9 x}{e ^{x}}

What is the integral from 0 to 2 of (e^x)/(1+e^x) ?
The integral from 0 to 2 of (e^x)/(1+e^x) is ln(1+e^2)-ln(2)