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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

= \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}

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

Substitute back

u = y - 1

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{\partial}{\partial x} ((x + x y + y)^{3})

\int x (3 x^{2} - 7)^{- 2} d x

\frac{d}{d x} (\frac{3}{4} x^{4} - 5 x^{2} + 4 x + 1)

\frac{d}{d x} (cos^{2} (2 - x^{2}))

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