What is the integral of 1/(1-2sqrt(y)) ?

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

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integral of 1/(1-2sqrt(y))

\int \frac{1}{1 - 2 y} d y

\frac{1}{2} - y - \frac{1}{2} ln ∣ 1 - 2 y ∣ + C

Solution steps

\int \frac{1}{1 - 2 y} d y

Apply Integration By Parts

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

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

\int \frac{y}{( 1 - 2 y ) ^{2}} d y = \frac{y}{1 - 2 y} - \frac{1}{2} + y + \frac{1}{2} ln ∣ 1 - 2 y ∣

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

Simplify

= \frac{1}{2} - y - \frac{1}{2} ln ∣ 1 - 2 y ∣

Add a constant to the solution

= \frac{1}{2} - y - \frac{1}{2} ln ∣ 1 - 2 y ∣ + C

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