What is the integral from 6 to 8 of (10)/((x-6)^3) ?

The integral from 6 to 8 of (10)/((x-6)^3) is diverges

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integral from 6 to 8 of (10)/((x-6)^3)

\int_{6}^{8} \frac{10}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{10}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 10 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 10 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 10 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

f (x) = x^{2} e^{- 7 x}

\int x^{2} \cdot 2 x d x

\int_{0}^{7} \frac{7 t}{( t - 8 ) ^{2}} d t

\frac{\partial}{\partial x} (sin^{2} (x) cos^{2} (y))

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What is the integral from 6 to 8 of (10)/((x-6)^3) ?
The integral from 6 to 8 of (10)/((x-6)^3) is diverges