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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\int_{2}^{4} g (x) d x

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

\int_{5}^{6} 6 x d x

\int_{a}^{b} x e^{x} d x

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

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