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

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

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integral from 0 to 1 of 1/(xsqrt(1-x))

\int_{0}^{1} \frac{1}{x 1 - x} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{x 1 - x} d x

Apply u-substitution

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

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

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

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

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}{- u ^{2} + 1} d u)

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - (- 2 [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1})

Simplify

- (- 2 [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1}) : 2 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{1}

= 2 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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