What is the integral of 1/(x^2)ln(x) ?

The integral of 1/(x^2)ln(x) is -(ln(x))/x-1/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x^2)ln(x)

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

- \frac{ln ( x )}{x} - \frac{1}{x} + C

Solution steps

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

Apply Integration By Parts

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

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

= - \frac{ln ( x )}{x} - \frac{1}{x}

Add a constant to the solution

= - \frac{ln ( x )}{x} - \frac{1}{x} + C

What is the integral of 1/(x^2)ln(x) ?
The integral of 1/(x^2)ln(x) is -(ln(x))/x-1/x+C