What is the integral from-9 to 0 of 1/((x-9)^2) ?

The integral from-9 to 0 of 1/((x-9)^2) is 1/18

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integral from-9 to 0 of 1/((x-9)^2)

\int_{- 9}^{0} \frac{1}{( x - 9 ) ^{2}} d x

\frac{1}{18}

Decimal Notation

0.05555 \dots

Solution steps

\int_{- 9}^{0} \frac{1}{( x - 9 ) ^{2}} d x

Apply u-substitution

: \int_{- 18}^{- 9} \frac{1}{u ^{2}} d u

= \int_{- 18}^{- 9} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{- 18}^{- 9}

= [- \frac{1}{u}]_{- 18}^{- 9}

Compute the boundaries:

\frac{1}{18}

= \frac{1}{18}

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What is the integral from-9 to 0 of 1/((x-9)^2) ?
The integral from-9 to 0 of 1/((x-9)^2) is 1/18