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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = ln (x)

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

Add a constant to the solution

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

\int x^{3} ln (x) d x

\int \frac{1}{25 + y ^{2}} d y

\int \frac{sin ( 2 t )}{cos ^{2} ( 2 t )} d t

\int tan^{6} (4 x) sec^{4} (4 x) d x

\int \frac{x - 2}{x + 2} d x

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