What is the integral from 1 to 2 of (x-1)^{-3} ?

The integral from 1 to 2 of (x-1)^{-3} is diverges

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integral from 1 to 2 of (x-1)^{-3}

\int_{1}^{2} (x - 1)^{- 3} d x

diverges

Solution steps

\int_{1}^{2} (x - 1)^{- 3} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int_{1}^{2} \frac{1}{( x - 1 ) ^{3}} d x

Apply u-substitution

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

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

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{0}^{1}

= [- \frac{1}{2 u ^{2}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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What is the integral from 1 to 2 of (x-1)^{-3} ?
The integral from 1 to 2 of (x-1)^{-3} is diverges