What is the integral from 6 to 8 of (36)/((x-6)^3) ?

The integral from 6 to 8 of (36)/((x-6)^3) is diverges

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integral from 6 to 8 of (36)/((x-6)^3)

\int_{6}^{8} \frac{36}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{36}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 36 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 36 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 36 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

2 x^{2} ln (2 x)

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

x \to 0 lim (- cos (\frac{1}{x}))

f (x) = 4 x^{2} + 2 x, at x = 2

\frac{\partial}{\partial x} (sech^{2} (x - t))

What is the integral from 6 to 8 of (36)/((x-6)^3) ?
The integral from 6 to 8 of (36)/((x-6)^3) is diverges