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

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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

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

Apply the Power Rule

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

= \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

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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