What is the integral from 2 to 9 of 1/(7(x-2)^3) ?

The integral from 2 to 9 of 1/(7(x-2)^3) is diverges

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integral from 2 to 9 of 1/(7(x-2)^3)

\int_{2}^{9} \frac{1}{7 ( x - 2 ) ^{3}} d x

diverges

Solution steps

\int_{2}^{9} \frac{1}{7 ( x - 2 ) ^{3}} d x

Take the constant out:

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

= \frac{1}{7} \cdot \int_{2}^{9} \frac{1}{( x - 2 ) ^{3}} d x

Apply u-substitution

: \int_{0}^{7} \frac{1}{u ^{3}} d u

= \frac{1}{7} \cdot \int_{0}^{7} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{0}^{7}

= \frac{1}{7} [- \frac{1}{2 u ^{2}}]_{0}^{7}

Compute the boundaries:

diverges

= diverges

x \to - 7 - lim (\frac{1}{x ^{2} ( x + 7 )})

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

\int_{5}^{53} 100 - x^{2} d x

\int e^{\frac{1}{2}}

cos^{2} (x) - sin^{2} (x)

What is the integral from 2 to 9 of 1/(7(x-2)^3) ?
The integral from 2 to 9 of 1/(7(x-2)^3) is diverges