What is the integral of 1/(x(1-x)^{1/2)} ?

The integral of 1/(x(1-x)^{1/2)} is -ln|(1-x)^{1/2}+1|+ln|(1-x)^{1/2}-1|+C

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integral of 1/(x(1-x)^{1/2)}

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

- ln (1 - x)^{\frac{1}{2}} + 1 + ln (1 - x)^{\frac{1}{2}} - 1 + C

Solution steps

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

Apply u-substitution

= \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

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 = (1 - x)^{\frac{1}{2}}

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of 1/(x(1-x)^{1/2)} ?
The integral of 1/(x(1-x)^{1/2)} is -ln|(1-x)^{1/2}+1|+ln|(1-x)^{1/2}-1|+C