What is the integral of 1/(sqrt(x)(1-x)) ?

The integral of 1/(sqrt(x)(1-x)) is ln|sqrt(x)+1|-ln|sqrt(x)-1|+C

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integral of 1/(sqrt(x)(1-x))

\int \frac{1}{x ( 1 - x )} d x

ln x + 1 - ln x - 1 + C

Solution steps

\int \frac{1}{x ( 1 - x )} d x

Factor

\frac{1}{x ( 1 - x )}

= \int \frac{1}{- x ( x - 1 )} d x

Take the constant out:

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

= - \int \frac{1}{x ( x - 1 )} d x

Apply u-substitution

: \int \frac{2}{u ^{2} - 1} d u

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

= - 2 \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= - 2 (- \int \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}))

Substitute back

u = x

= - 2 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}))

Simplify

- 2 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})) : ln x + 1 - ln x - 1

= ln x + 1 - ln x - 1

Add a constant to the solution

= ln x + 1 - ln x - 1 + C

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What is the integral of 1/(sqrt(x)(1-x)) ?
The integral of 1/(sqrt(x)(1-x)) is ln|sqrt(x)+1|-ln|sqrt(x)-1|+C