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

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

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integral of 1/(y-y^2)

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

ln ∣ y ∣ - ln ∣ y - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{y - y ^{2}} : \frac{1}{y} - \frac{1}{y - 1}

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

Apply the Sum Rule:

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

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

\int \frac{1}{y} d y = ln ∣ y ∣

\int \frac{1}{y - 1} d y = ln ∣ y - 1 ∣

= ln ∣ y ∣ - ln ∣ y - 1 ∣

Add a constant to the solution

= ln ∣ y ∣ - ln ∣ y - 1 ∣ + C

\frac{\partial}{\partial x} (18 x y e^{- x^{2} - y^{2}})

\frac{\partial}{\partial x} (s t)

x = - 2, x = 3, y = 2 x, y = x^{2} - 3

\frac{\partial}{\partial x} (lo g_{e} (x + y))

f (x) = 5^{x}

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