What is the integral from 6 to 8 of (72)/((x-6)^3) ?

The integral from 6 to 8 of (72)/((x-6)^3) is diverges

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integral from 6 to 8 of (72)/((x-6)^3)

\int_{6}^{8} \frac{72}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{72}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 72 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 72 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 72 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

x \to 3 lim (f (x))

\frac{d y}{d x} = \frac{x ^{2} y ^{2}}{1 + 2}

\int \frac{7 - 98 x}{9 - 49 x ^{2}} d x

y = 5 + 4 x^{2} - 2 x^{3}

x^{2} + 4 x - 3

What is the integral from 6 to 8 of (72)/((x-6)^3) ?
The integral from 6 to 8 of (72)/((x-6)^3) is diverges