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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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

\int X d X

\int \frac{x}{2 x ^{2} + 3} d x

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

- x^{2} + 3 x, 0

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