What is the integral from 1 to 2 of 1/(x(ln(x))^p) ?

The integral from 1 to 2 of 1/(x(ln(x))^p) is (ln^{-p+1}(2))/(-p+1)

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integral from 1 to 2 of 1/(x(ln(x))^p)

\int_{1}^{2} \frac{1}{x ( ln ( x ) ) ^{p}} d x

\frac{ln ^{- p + 1} ( 2 )}{- p + 1}

Solution steps

\int_{1}^{2} \frac{1}{x ( ln ( x ) ) ^{p}} d x

Apply exponent rule:

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

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

= \int_{1}^{2} (ln (x))^{- p} \frac{1}{x} d x

Simplify

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

= \int_{1}^{2} \frac{ln ^{- p} ( x )}{x} d x

Apply u-substitution

: \int_{0}^{l n (2)} u^{- p} d u

= \int_{0}^{l n (2)} u^{- p} d u

Apply the Power Rule

: [\frac{u ^{- p + 1}}{- p + 1}]_{0}^{l n (2)}

= [\frac{u ^{- p + 1}}{- p + 1}]_{0}^{l n (2)}

Compute the boundaries:

\frac{ln ^{- p + 1} ( 2 )}{- p + 1}

= \frac{ln ^{- p + 1} ( 2 )}{- p + 1}

f (x) = ln (3 x e^{- x})

\frac{f ( x )}{x ^{5}}

\frac{\partial}{\partial x} (sin (x^{2} + y^{3}))

\int - \frac{1}{4 - x} d x

(x^{2} + 4) sin (5 x)

What is the integral from 1 to 2 of 1/(x(ln(x))^p) ?
The integral from 1 to 2 of 1/(x(ln(x))^p) is (ln^{-p+1}(2))/(-p+1)