What is the integral from 6 to 8 of (26)/((x-6)^3) ?

The integral from 6 to 8 of (26)/((x-6)^3) is diverges

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integral from 6 to 8 of (26)/((x-6)^3)

\int_{6}^{8} \frac{26}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{26}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 26 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 26 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 26 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\frac{\partial}{\partial x} (- 4 x y (3 - x - y))

y^{^{''}} + 2 y^{^{'}} + 37 y = 5 e^{- t} cos (6 t)

\int e^{a x} sin (0 x) d x

f (x) = 8 x - 3

x = 9 y^{2}, x = 4 + 5 y^{2}

What is the integral from 6 to 8 of (26)/((x-6)^3) ?
The integral from 6 to 8 of (26)/((x-6)^3) is diverges