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

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

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

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

diverges

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 8 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 8 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

x \to - 7 lim (2 x^{2} - 6 x)

(x + 9) (7 x + 2)

f (x) = 3 x, (3, 3)

\frac{2 s + 2}{s ^{2} + 4 s + 5}

\int \frac{1}{2 + 3 x ^{2}} d x

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