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

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

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

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

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

Solution steps

\int \frac{1}{x 1 - x} 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

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{1 + x ^{4}} d x

\int_{0}^{2} x^{5} e^{x^{3}} d x

4 x^{2} - 3, - x^{2} + 9

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 4 x + 4 y)

f (x) = ln (x), at x = 2

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