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

The integral of (sqrt(x))/(1-x) is ln|sqrt(x)+1|-ln|sqrt(x)-1|-2sqrt(x)+C

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

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

ln x + 1 - ln x - 1 - 2 x + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d u = u

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

Substitute back

u = x

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

Simplify

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

= ln x + 1 - ln x - 1 - 2 x

Add a constant to the solution

= ln x + 1 - ln x - 1 - 2 x + C

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