What is the integral from a to b of 1/((x-a)^n) ?

The integral from a to b of 1/((x-a)^n) is ((b-a)^{-n+1})/(-n+1)

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integral from a to b of 1/((x-a)^n)

\int_{a}^{b} \frac{1}{( x - a ) ^{n}} d x

\frac{( b - a ) ^{- n + 1}}{- n + 1}

Solution steps

\int_{a}^{b} \frac{1}{( x - a ) ^{n}} d x

Apply u-substitution

: \int_{0}^{b - a} \frac{1}{u ^{n}} d u

= \int_{0}^{b - a} \frac{1}{u ^{n}} d u

Apply the Power Rule

: [\frac{u ^{- n + 1}}{- n + 1}]_{0}^{b - a}

= [\frac{u ^{- n + 1}}{- n + 1}]_{0}^{b - a}

Compute the boundaries:

\frac{( b - a ) ^{- n + 1}}{- n + 1}

= \frac{( b - a ) ^{- n + 1}}{- n + 1}

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What is the integral from a to b of 1/((x-a)^n) ?
The integral from a to b of 1/((x-a)^n) is ((b-a)^{-n+1})/(-n+1)