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

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

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

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int ln^{- p} (x) \frac{1}{x} d x

Simplify

ln^{- p} (x) \frac{1}{x} : \frac{ln ^{- p} ( x )}{x}

= \int \frac{ln ^{- p} ( x )}{x} d x

Apply u-substitution

: \int u^{- p} d u

= \int u^{- p} d u

Apply the Power Rule

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

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

Substitute back

u = ln (x)

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

Add a constant to the solution

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

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