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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (ln(z))/(z^2)

\int \frac{ln ( z )}{z ^{2}} d z

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

Solution steps

\int \frac{ln ( z )}{z ^{2}} d z

Apply Integration By Parts

: - \frac{ln ( z )}{z} - \int - \frac{1}{z ^{2}} d z

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

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

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

Add a constant to the solution

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

\int_{- 1}^{3} \frac{4 x ^{2} - 7}{2 x + 3} d x

\frac{d}{d x} (I n)

\frac{d y}{d x} = 12 x^{3} y

\int (e^{x} + 1) (e^{x} + x) d x

x^{4} (2 x - 5)^{6}

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