What is the integral from 3 to 4 of-1/((x-3)^3) ?

The integral from 3 to 4 of-1/((x-3)^3) is diverges

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integral from 3 to 4 of-1/((x-3)^3)

\int_{3}^{4} - \frac{1}{( x - 3 ) ^{3}} d x

diverges

Solution steps

\int_{3}^{4} - \frac{1}{( x - 3 ) ^{3}} d x

Take the constant out:

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

= - \int_{3}^{4} \frac{1}{( x - 3 ) ^{3}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

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x, - 1, 1

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What is the integral from 3 to 4 of-1/((x-3)^3) ?
The integral from 3 to 4 of-1/((x-3)^3) is diverges