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

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

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

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

diverges

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 6 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 6 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

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

(sin (x) + 2 tan (x)) (sec (x) - 3 csc (x))

\frac{d}{d x} (\frac{1}{x ^{12}})

tan (2 x^{2} - x + 3)

y = 9 sec (\frac{1}{3} x)

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