What is the integral from-8 to 0 of 1/((x-8)^2) ?

The integral from-8 to 0 of 1/((x-8)^2) is 1/16

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integral from-8 to 0 of 1/((x-8)^2)

\int_{- 8}^{0} \frac{1}{( x - 8 ) ^{2}} d x

\frac{1}{16}

Decimal Notation

0.0625

Solution steps

\int_{- 8}^{0} \frac{1}{( x - 8 ) ^{2}} d x

Apply u-substitution

: \int_{- 16}^{- 8} \frac{1}{u ^{2}} d u

= \int_{- 16}^{- 8} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{- 16}^{- 8}

= [- \frac{1}{u}]_{- 16}^{- 8}

Compute the boundaries:

\frac{1}{16}

= \frac{1}{16}

\frac{d y}{d x} = \frac{( y ^{2} + 4 )}{2 x y}

\frac{d y}{d x} = \frac{cos ( 8 x )}{cos ( 4 y )}

\frac{s + 3}{s + 5}

\int e^{i 7 x} cos (3 x) d x

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

What is the integral from-8 to 0 of 1/((x-8)^2) ?
The integral from-8 to 0 of 1/((x-8)^2) is 1/16