What is the integral of 1/(zln(z)) ?

The integral of 1/(zln(z)) is ln|ln(z)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(zln(z))

\int \frac{1}{z ln ( z )} d z

ln ∣ ln (z) ∣ + C

Solution steps

\int \frac{1}{z ln ( z )} d z

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = ln (z)

= ln ∣ ln (z) ∣

Add a constant to the solution

= ln ∣ ln (z) ∣ + C

\int \frac{x ^{2} + 4 x + 1}{( x ^{2} - x + 1 ) ( x + 4 )} d x

\int (\frac{1}{4})^{3 x} d x

\int \frac{3 x ^{2} - 5 x}{x} d x

e^{- 4 x}

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