What is the integral from 6 to 8 of (20)/((x-6)^3) ?

The integral from 6 to 8 of (20)/((x-6)^3) is diverges

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integral from 6 to 8 of (20)/((x-6)^3)

\int_{6}^{8} \frac{20}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{20}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 20 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 20 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 20 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (f (x) (e^{x}))

y^{^{'}} - 6 y^{^{'}} + 9 y = 0

n = 0 \sum \infty 3 (- \frac{1}{3})^{n}

y = x^{3} - 2 x + 1, (4, 57)

f (x) = \frac{sin ( x )}{1 + cos ( x )}

What is the integral from 6 to 8 of (20)/((x-6)^3) ?
The integral from 6 to 8 of (20)/((x-6)^3) is diverges