What is the integral from 6 to 8 of (90)/((x-6)^3) ?

The integral from 6 to 8 of (90)/((x-6)^3) is diverges

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integral from 6 to 8 of (90)/((x-6)^3)

\int_{6}^{8} \frac{90}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{90}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 90 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 90 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 90 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

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\int (ln (x))^{5} d x

t^{3} \frac{d y}{d t} + 3 t^{2} y = 8 cos (t), y (π) = 0

y = (x^{2} - 2 x + 2) e^{\frac{5 x}{2}}

\int sec (θ) tan (θ) d θ

What is the integral from 6 to 8 of (90)/((x-6)^3) ?
The integral from 6 to 8 of (90)/((x-6)^3) is diverges