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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\frac{1 + x}{1 - x}

\int_{0}^{3} π (2 x)^{2} d x

\int \frac{1}{( 1 - y ^{2} ) ^{\frac{1}{2}}} d y

\int \frac{x + 1}{x ^{2} + 2 x + 2} d x

\int \frac{x ^{2} - 4}{x ^{3} - 2 x ^{2} + x} d x

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