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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{1}{x}} x y d y

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

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

\int_{0}^{5} (25 - x^{2})^{\frac{1}{2}} d x

\int_{- 1}^{0} (2 + 1 - x^{2}) d x

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