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

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

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

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

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

Solution steps

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

Apply Integration By Parts

: [\frac{1}{- p + 1} (x^{- p + 1} ln (x) - \int \frac{1}{x ^{p} ( - p + 1 )} d x (- p + 1))]_{1}^{e}

= [\frac{1}{- p + 1} (x^{- p + 1} ln (x) - \int \frac{1}{x ^{p} ( - p + 1 )} d x (- p + 1))]_{1}^{e}

\int \frac{1}{x ^{p} ( - p + 1 )} d x = \frac{x ^{- p + 1}}{( - p + 1 ) ^{2}}

= [\frac{1}{- p + 1} (x^{- p + 1} ln (x) - \frac{x ^{- p + 1}}{( - p + 1 ) ^{2}} (- p + 1))]_{1}^{e}

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

= [\frac{1}{- p + 1} (x^{- p + 1} ln (x) - \frac{x ^{- p + 1}}{- p + 1})]_{1}^{e}

Compute the boundaries:

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

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

Simplify

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

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