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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\int - \frac{y}{( x + 2 ) ^{2}} d x

f (x) = x^{3} + 3 x, at x = 2

\frac{\partial}{\partial x} (x \cdot cos (y))

x \to \infty lim ((x)^{3})

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