What is the integral of 1/(2x(3-x)) ?

The integral of 1/(2x(3-x)) is 1/2 (1/3 ln|x|-1/3 ln|x-3|)+C

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integral of 1/(2x(3-x))

\int \frac{1}{2 x ( 3 - x )} d x

\frac{1}{2} (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 3 ∣) + C

Solution steps

\int \frac{1}{2 x ( 3 - x )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{x ( 3 - x )} d x

Take the partial fraction of

\frac{1}{x ( 3 - x )} : \frac{1}{3 x} - \frac{1}{3 ( x - 3 )}

= \frac{1}{2} \cdot \int \frac{1}{3 x} - \frac{1}{3 ( x - 3 )} 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

= \frac{1}{2} (\int \frac{1}{3 x} d x - \int \frac{1}{3 ( x - 3 )} d x)

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

\int \frac{1}{3 ( x - 3 )} d x = \frac{1}{3} ln ∣ x - 3 ∣

= \frac{1}{2} (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 3 ∣)

Add a constant to the solution

= \frac{1}{2} (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 3 ∣) + C

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What is the integral of 1/(2x(3-x)) ?
The integral of 1/(2x(3-x)) is 1/2 (1/3 ln|x|-1/3 ln|x-3|)+C