What is the integral of zln(z) ?

The integral of zln(z) is 1/2 z^2ln(z)-(z^2)/4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of zln(z)

\int z ln (z) d z

\frac{1}{2} z^{2} ln (z) - \frac{z ^{2}}{4} + C

Solution steps

\int z ln (z) d z

Apply Integration By Parts

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

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

\int \frac{z}{2} d z = \frac{z ^{2}}{4}

= \frac{1}{2} z^{2} ln (z) - \frac{z ^{2}}{4}

Add a constant to the solution

= \frac{1}{2} z^{2} ln (z) - \frac{z ^{2}}{4} + C

\int x^{- 1} \cdot \frac{1}{2 + x ^{2}} d x

\int (2 x + 1) \cdot ln (x) d x

\int \frac{( x + 3 ) ^{7}}{2 x} d x

\int 4 y + 2 t - 5 d t

\int (sec (x))^{2} (tan (x))^{2} d x

What is the integral of zln(z) ?
The integral of zln(z) is 1/2 z^2ln(z)-(z^2)/4+C