What is the integral from 2 to 5 of 1/((x-2)^n) ?

The integral from 2 to 5 of 1/((x-2)^n) is (3^{-n+1})/(-n+1)

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integral from 2 to 5 of 1/((x-2)^n)

\int_{2}^{5} \frac{1}{( x - 2 ) ^{n}} d x

\frac{3 ^{- n + 1}}{- n + 1}

Solution steps

\int_{2}^{5} \frac{1}{( x - 2 ) ^{n}} d x

Apply u-substitution

: \int_{0}^{3} \frac{1}{u ^{n}} d u

= \int_{0}^{3} \frac{1}{u ^{n}} d u

Apply the Power Rule

: [\frac{u ^{- n + 1}}{- n + 1}]_{0}^{3}

= [\frac{u ^{- n + 1}}{- n + 1}]_{0}^{3}

Compute the boundaries:

\frac{3 ^{- n + 1}}{- n + 1}

= \frac{3 ^{- n + 1}}{- n + 1}

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\int_{1}^{4} 5 x x + 1 d x

What is the integral from 2 to 5 of 1/((x-2)^n) ?
The integral from 2 to 5 of 1/((x-2)^n) is (3^{-n+1})/(-n+1)