What is the integral from 6 to 8 of (92)/((x-6)^3) ?

The integral from 6 to 8 of (92)/((x-6)^3) is diverges

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integral from 6 to 8 of (92)/((x-6)^3)

\int_{6}^{8} \frac{92}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{92}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 92 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 92 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 92 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (sin^{4} (cos (8 x)))

x \to 4 lim (x^{4} - 5 x^{2} + 2)

f (x) = \frac{1}{x}, [1, 2]

\int_{- 5}^{0} 2^{1 - x} d x

y = \frac{6 x}{x ^{2} + 1}

What is the integral from 6 to 8 of (92)/((x-6)^3) ?
The integral from 6 to 8 of (92)/((x-6)^3) is diverges