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

The integral from 0 to 1 of 1/(x^p) is 1/(-p+1)

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

\int_{0}^{1} \frac{1}{x ^{p}} d x

\frac{1}{- p + 1}

Solution steps

\int_{0}^{1} \frac{1}{x ^{p}} d x

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{- p + 1}

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

\frac{\partial}{\partial t} (- (s^{2} + t^{2}))

\frac{d}{d x} ((4 x - 5)^{3})

\int \frac{1}{e ^{- 3 x}} d x

\frac{d}{d y} (cos (x) y)

\int \frac{x ^{3} - 1}{x ^{3} + 2 x ^{2} + x} d x

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