What is the integral from 3 to 4 of 1/((x-4)^2) ?

The integral from 3 to 4 of 1/((x-4)^2) is diverges

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integral from 3 to 4 of 1/((x-4)^2)

\int_{3}^{4} \frac{1}{( x - 4 ) ^{2}} d x

diverges

Solution steps

\int_{3}^{4} \frac{1}{( x - 4 ) ^{2}} d x

Apply u-substitution

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

= \int_{- 1}^{0} \frac{1}{u ^{2}} d u

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

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What is the integral from 3 to 4 of 1/((x-4)^2) ?
The integral from 3 to 4 of 1/((x-4)^2) is diverges