What is the integral from 6 to 8 of (30)/((x-6)^3) ?

The integral from 6 to 8 of (30)/((x-6)^3) is diverges

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integral from 6 to 8 of (30)/((x-6)^3)

\int_{6}^{8} \frac{30}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{30}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 30 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 30 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 30 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\int_{0}^{x} (x + 2 y) d y

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

\int x sin (\frac{1}{3}) x d x

t y^{^{'}} + 4 y = t^{2} - t + 5

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

What is the integral from 6 to 8 of (30)/((x-6)^3) ?
The integral from 6 to 8 of (30)/((x-6)^3) is diverges