What is the integral from 0 to 2 of 1/((2-x)^2) ?

The integral from 0 to 2 of 1/((2-x)^2) is diverges

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integral from 0 to 2 of 1/((2-x)^2)

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

diverges

Solution steps

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

diverges

= diverges

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What is the integral from 0 to 2 of 1/((2-x)^2) ?
The integral from 0 to 2 of 1/((2-x)^2) is diverges