What is the integral from 1 to 8 of 1/(x^p) ?

The integral from 1 to 8 of 1/(x^p) is (8^{-p+1}-1)/(-p+1)

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integral from 1 to 8 of 1/(x^p)

\int_{1}^{8} \frac{1}{x ^{p}} d x

\frac{8 ^{- p + 1} - 1}{- p + 1}

Solution steps

\int_{1}^{8} \frac{1}{x ^{p}} d x

Apply the Power Rule

: [\frac{x ^{- p + 1}}{- p + 1}]_{1}^{8}

= [\frac{x ^{- p + 1}}{- p + 1}]_{1}^{8}

Compute the boundaries:

\frac{8 ^{- p + 1}}{- p + 1} - \frac{1}{- p + 1}

= \frac{8 ^{- p + 1}}{- p + 1} - \frac{1}{- p + 1}

Simplify

= \frac{8 ^{- p + 1} - 1}{- p + 1}

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What is the integral from 1 to 8 of 1/(x^p) ?
The integral from 1 to 8 of 1/(x^p) is (8^{-p+1}-1)/(-p+1)