What is the integral from 6 to 8 of (98)/((x-6)^3) ?

The integral from 6 to 8 of (98)/((x-6)^3) is diverges

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integral from 6 to 8 of (98)/((x-6)^3)

\int_{6}^{8} \frac{98}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{98}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 98 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 98 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 98 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (7 x^{3} - 5 x + 8)

x \to 2 lim (- 1.999 + 2)

f (x) = - x^{2} + 4 x

\int \frac{10 x - 2}{5 x ^{2} - 2 x} d x

\int \frac{( 1 + sin ( x ) )}{cos ^{2} ( x )} d x

What is the integral from 6 to 8 of (98)/((x-6)^3) ?
The integral from 6 to 8 of (98)/((x-6)^3) is diverges