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

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

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

ln (e + 2)

Decimal Notation

1.55144 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e + 2}

Compute the boundaries:

ln (e + 2)

= ln (e + 2)

\int_{\frac{π}{6}}^{\frac{π}{3}} 5 csc^{3} (x) d x

\int_{0}^{π} \frac{1}{cos ( x ) - 1} d x

\int_{0}^{3} \frac{1}{x ^{2} + 1} d x

\int_{- 1}^{4} 5 x^{4} - 8 x^{3} + 6 d x

\int_{0}^{x} cosh^{2} (t) d t

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