What is the integral from 6 to 8 of 4/((x-6)^3) ?

The integral from 6 to 8 of 4/((x-6)^3) is diverges

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integral from 6 to 8 of 4/((x-6)^3)

\int_{6}^{8} \frac{4}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{4}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 4 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 4 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 4 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\frac{d y}{d x} = \frac{x - 5}{y ^{2}}

\int (\frac{2 x ^{3}}{x ^{2} - 4}) d x

\frac{d}{d x} (x^{2} cos (x) + 2 x sin (x))

- 36 sin (4 x)

s^{2} + 4 2 t + 1 + \frac{s}{r} + \frac{t ^{6} r ^{2}}{t + r ^{3}}

What is the integral from 6 to 8 of 4/((x-6)^3) ?
The integral from 6 to 8 of 4/((x-6)^3) is diverges