What is the integral from 1/e to 1 of ln(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1/e to 1 of ln(x)

\int_{\frac{1}{e}}^{1} ln (x) d x

- 1 + \frac{2}{e}

Decimal Notation

- 0.26424 \dots

Solution steps

\int_{\frac{1}{e}}^{1} ln (x) d x

Apply Integration By Parts

: [x ln (x) - \int 1 d x]_{\frac{1}{e}}^{1}

= [x ln (x) - \int 1 d x]_{\frac{1}{e}}^{1}

\int 1 d x = x

= [x ln (x) - x]_{\frac{1}{e}}^{1}

Compute the boundaries:

- 1 + \frac{2}{e}

= - 1 + \frac{2}{e}

\int_{0}^{2} - x^{3} + 4 x d x

\int_{1}^{9} (x^{2} + 2) d x

\int_{2}^{7} \frac{16 x}{x ^{2} + 1} d x

\int_{0}^{2} (6 x^{2} - 3 x^{3}) d x

\int_{0}^{2} \frac{2 x}{( x ^{2} + 4 ) ^{3}} d x

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