What is the integral from 6 to 8 of (60)/((x-6)^3) ?

The integral from 6 to 8 of (60)/((x-6)^3) is diverges

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integral from 6 to 8 of (60)/((x-6)^3)

\int_{6}^{8} \frac{60}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{60}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 60 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 60 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 60 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

y = ln (x), at x = 2

\frac{s}{( s ^{2} + 1 ^{2} ) ^{2}}

\int (1 - csc (t) cot (t)) d t

\frac{d}{d x} (a e^{2 x} + b x e^{2 x})

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}} + 3 x)

What is the integral from 6 to 8 of (60)/((x-6)^3) ?
The integral from 6 to 8 of (60)/((x-6)^3) is diverges