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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

diverges

= diverges

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

\int_{0}^{\frac{3 π}{2}} x sin (x) d x

\int_{1}^{7} ∣ sin (x) ∣ d x

\int_{0}^{1} (1 - x) sin (nπ x) d x

\int_{\frac{π}{2}}^{\frac{3 π}{2}} x sin (x) d x

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