What is the integral of 1/(y(4-y)) ?

The integral of 1/(y(4-y)) is 1/4 ln|y|-1/4 ln|y-4|+C

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integral of 1/(y(4-y))

\int \frac{1}{y ( 4 - y )} d y

\frac{1}{4} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 4 ∣ + C

Solution steps

\int \frac{1}{y ( 4 - y )} d y

Take the partial fraction of

\frac{1}{y ( 4 - y )} : \frac{1}{4 y} - \frac{1}{4 ( y - 4 )}

= \int \frac{1}{4 y} - \frac{1}{4 ( y - 4 )} 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}{4 y} d y - \int \frac{1}{4 ( y - 4 )} d y

\int \frac{1}{4 y} d y = \frac{1}{4} ln ∣ y ∣

\int \frac{1}{4 ( y - 4 )} d y = \frac{1}{4} ln ∣ y - 4 ∣

= \frac{1}{4} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 4 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 4 ∣ + C

\frac{d}{d x} (7 (e^{x} cos (x) - e^{x} sin (x)))

\frac{d y}{d x} = \frac{3 x}{5 y}

x^{2} + 4

\frac{\partial}{\partial x} (y^{x z} + 2 (yz)^{x})

What is the integral of 1/(y(4-y)) ?
The integral of 1/(y(4-y)) is 1/4 ln|y|-1/4 ln|y-4|+C