What is the integral from 6 to 8 of (46)/((x-6)^3) ?

The integral from 6 to 8 of (46)/((x-6)^3) is diverges

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integral from 6 to 8 of (46)/((x-6)^3)

\int_{6}^{8} \frac{46}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{46}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 46 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 46 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 46 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

f (x) = \frac{6 x}{x ^{2} - 3}, (3, 3)

\frac{d y}{d x} = - 4 y + x^{3} - x

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

\int \frac{a + b x ^{6}}{7 a x + b x ^{7}} d x

4 e^{- 5 t} - 3 e^{- t}

What is the integral from 6 to 8 of (46)/((x-6)^3) ?
The integral from 6 to 8 of (46)/((x-6)^3) is diverges