What is the integral from 1 to 0 of (ln(x))/x ?

The integral from 1 to 0 of (ln(x))/x is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 0 of (ln(x))/x

\int_{1}^{0} \frac{ln ( x )}{x} d x

diverges

Solution steps

\int_{1}^{0} \frac{ln ( x )}{x} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} \frac{ln ( x )}{x} d x

Apply Integration By Parts

: [ln (x) - \int \frac{1 - ln ( x )}{x} d x]_{0}^{1}

= - [ln (x) - \int \frac{1 - ln ( x )}{x} d x]_{0}^{1}

\int \frac{1 - ln ( x )}{x} d x = - \frac{( 1 - ln ( x ) ) ^{2}}{2}

= - [ln (x) - (- \frac{( 1 - ln ( x ) ) ^{2}}{2})]_{0}^{1}

Simplify

- [ln (x) - (- \frac{( 1 - ln ( x ) ) ^{2}}{2})]_{0}^{1} : - [\frac{1}{2} (ln^{2} (x) + 1)]_{0}^{1}

= - [\frac{1}{2} (ln^{2} (x) + 1)]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int_{0}^{16} x^{2} d x

\int_{- 1}^{0} 3 x + 3 d x

\int_{- 1}^{2} ((x + 3) - (x^{2} + 1)) d x

\int_{0}^{π} 93 sin (x) d x

\int_{0}^{4} 42 t d t

What is the integral from 1 to 0 of (ln(x))/x ?
The integral from 1 to 0 of (ln(x))/x is diverges