What is the integral of 1/(xln^n(x)) ?

The integral of 1/(xln^n(x)) is (ln^{-n+1}(x))/(-n+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(xln^n(x))

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

= \frac{u ^{- n + 1}}{- n + 1}

Substitute back

u = ln (x)

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

Add a constant to the solution

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

What is the integral of 1/(xln^n(x)) ?
The integral of 1/(xln^n(x)) is (ln^{-n+1}(x))/(-n+1)+C