What is the integral from 6 to 8 of (24)/((x-6)^3) ?

The integral from 6 to 8 of (24)/((x-6)^3) is diverges

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integral from 6 to 8 of (24)/((x-6)^3)

\int_{6}^{8} \frac{24}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{24}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 24 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 24 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 24 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

f (x) = 5 x, (25, 25)

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 9 y = 0

y^{^{'}} + \frac{1}{165} y = 0.48

\frac{d}{d x} (10 sin (2 x) + 15 cos (2 x) + 100)

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

What is the integral from 6 to 8 of (24)/((x-6)^3) ?
The integral from 6 to 8 of (24)/((x-6)^3) is diverges