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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

1

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e}

Compute the boundaries:

1

= 1

\int cos^{3} (4 x) sin (4 x) d x

\frac{d y}{d x} + tan (x) y - sec (x) = 0

\frac{d}{d x} (26 - sec^{3} (4 x))

x \to \infty lim ((4^{x} + 8^{x})^{\frac{1}{2 \cdot x}})

\frac{d}{d x} (- x cos (x) + x)

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