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

The integral from 2 to 5 of 1/((x-2)^{-1)} is 9/2

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

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

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

Simplify

\frac{1}{( x - 2 ) ^{- 1}} : x - 2

= \int_{2}^{5} x - 2 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{5} x d x - \int_{2}^{5} 2 d x

\int_{2}^{5} x d x = \frac{21}{2}

\int_{2}^{5} 2 d x = 6

= \frac{21}{2} - 6

Simplify

\frac{21}{2} - 6 : \frac{9}{2}

= \frac{9}{2}

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What is the integral from 2 to 5 of 1/((x-2)^{-1)} ?
The integral from 2 to 5 of 1/((x-2)^{-1)} is 9/2