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

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

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

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

diverges

Solution steps

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

y = 8 + 4 x^{2} - 2 x^{3}

6 (2 e^{x^{2}} x^{2} + e^{x^{2}})

3 + x

x \to + 0 + lim (ln (2 x))

\int_{1}^{2} \frac{x}{2} d x

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