What is the integral from 6 to 8 of (82)/((x-6)^3) ?

The integral from 6 to 8 of (82)/((x-6)^3) is diverges

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integral from 6 to 8 of (82)/((x-6)^3)

\int_{6}^{8} \frac{82}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{82}{( x - 6 ) ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 82 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{u ^{3}} d u

= 82 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 82 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

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

\frac{\partial}{\partial x} (y^{2} cos (x y^{2}))

\int \frac{x ^{4}}{( x ^{4} - 1 )} d x

\frac{d}{d t} (e^{- 3 t} sin (3 t))

y = x^{\frac{7}{5}} + e^{- 4 x}

What is the integral from 6 to 8 of (82)/((x-6)^3) ?
The integral from 6 to 8 of (82)/((x-6)^3) is diverges