What is the integral from 0 to 1 of (e^x+6)/(e^x+6x) ?

The integral from 0 to 1 of (e^x+6)/(e^x+6x) is ln(e+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 1 of (e^x+6)/(e^x+6x)

\int_{0}^{1} \frac{e ^{x} + 6}{e ^{x} + 6 x} d x

ln (e + 6)

Solution steps

\int_{0}^{1} \frac{e ^{x} + 6}{e ^{x} + 6 x} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e + 6}

Compute the boundaries:

ln (e + 6)

= ln (e + 6)

\int_{- 4}^{4} 3 - ∣ x ∣ d x

\int_{0}^{2} (x^{2} + 2) d x

\int_{- 1}^{1} sin^{7} (x) d x

\int_{0}^{7} π (2 3 y)^{2} d y

\int_{0}^{1} 3 x e^{x} d x

What is the integral from 0 to 1 of (e^x+6)/(e^x+6x) ?
The integral from 0 to 1 of (e^x+6)/(e^x+6x) is ln(e+6)