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

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

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

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

diverges

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\int_{0}^{1} 16 x^{2} + 1 d x

\int_{x}^{y} 1 d x

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

\int_{0}^{\frac{1}{2}} e^{2 x} d x

\int_{- 2}^{2} (x - 5) d x

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