What is the integral from 6 to 8 of 1/((x-6)^3) ?

The integral from 6 to 8 of 1/((x-6)^3) is diverges

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integral from 6 to 8 of 1/((x-6)^3)

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

diverges

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\frac{1}{1 - 2 x}

\frac{d}{d x} (\frac{x}{2.4 ( c ^{2} + c x )} (c^{2} - x^{2}))

\int \frac{cos ( 2 t )}{( 1 + sin ( 2 t ) ) ^{4}} d t

x \to - 1 lim (\frac{x + 2}{x ^{3} + 1})

\int x x (1 + x^{2} x)^{4} d x

What is the integral from 6 to 8 of 1/((x-6)^3) ?
The integral from 6 to 8 of 1/((x-6)^3) is diverges