What is the integral from 6 to 8 of (14)/((x-6)^3) ?

The integral from 6 to 8 of (14)/((x-6)^3) is diverges

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integral from 6 to 8 of (14)/((x-6)^3)

\int_{6}^{8} \frac{14}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{14}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 14 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 14 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 14 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

x \to a lim ((\frac{sin ( x )}{sin ( a )})^{\frac{1}{x + a}})

\int \frac{2 x + 1}{3 x ^{2} + 3 x + 4} d x

3 x + 3, 4 x + 4, - 1, 3

\frac{\partial}{\partial x} (ln (2 x y + 4))

\frac{\partial}{\partial y} (arctan (x + 6 y))

What is the integral from 6 to 8 of (14)/((x-6)^3) ?
The integral from 6 to 8 of (14)/((x-6)^3) is diverges