What is the integral from 0 to 1 of 1/(x^pi) ?

The integral from 0 to 1 of 1/(x^pi) is (0^{-pi+1}-1)/(pi-1)

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integral from 0 to 1 of 1/(x^pi)

\int_{0}^{1} \frac{1}{x ^{π}} d x

\frac{0 ^{- π + 1} - 1}{π - 1}

Solution steps

\int_{0}^{1} \frac{1}{x ^{π}} d x

Apply the Power Rule

: [\frac{x ^{- π + 1}}{- π + 1}]_{0}^{1}

= [\frac{x ^{- π + 1}}{- π + 1}]_{0}^{1}

Compute the boundaries:

\frac{0 ^{- π + 1} - 1}{π - 1}

= \frac{0 ^{- π + 1} - 1}{π - 1}

\int_{0}^{1} \frac{( 1 - x ) ^{2}}{4} d x

\int_{0}^{π} (x + 1) d x

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\int_{1}^{e} (\frac{ln ( u ) + 1}{u}) d u

\int_{3}^{3.5} 2 x d x

What is the integral from 0 to 1 of 1/(x^pi) ?
The integral from 0 to 1 of 1/(x^pi) is (0^{-pi+1}-1)/(pi-1)