What is the integral from 6 to 8 of (48)/((x-6)^3) ?

The integral from 6 to 8 of (48)/((x-6)^3) is diverges

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integral from 6 to 8 of (48)/((x-6)^3)

\int_{6}^{8} \frac{48}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{48}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 48 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 48 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 48 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\frac{\partial}{\partial x} (\frac{x}{y + 8 z})

\int_{1}^{\infty} 6 e^{- 3 x} d x

u^{\frac{1}{2}}

\int_{0}^{4} 3 e^{- 2 x} d x

\frac{\partial}{\partial x} (x^{1} x^{2} + 10 x^{1})

What is the integral from 6 to 8 of (48)/((x-6)^3) ?
The integral from 6 to 8 of (48)/((x-6)^3) is diverges