What is the integral from 6 to 8 of (34)/((x-6)^3) ?

The integral from 6 to 8 of (34)/((x-6)^3) is diverges

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integral from 6 to 8 of (34)/((x-6)^3)

\int_{6}^{8} \frac{34}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{34}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 34 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 34 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 34 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\frac{\partial}{\partial x} (x y (\frac{( 2 - x y )}{x + y}))

\frac{\partial ^{2}}{\partial y \partial x} (\frac{x y}{x - y})

\int \frac{x ^{2} - 36}{x ^{2}} d x

\int \frac{1}{16 + x ^{2}} d x

x \to 0 + lim (- \frac{3 ln ( x )}{x})

What is the integral from 6 to 8 of (34)/((x-6)^3) ?
The integral from 6 to 8 of (34)/((x-6)^3) is diverges