What is the integral of 1/(x(4-x^2)) ?

The integral of 1/(x(4-x^2)) is 1/4 ln|x|-1/8 ln|x+2|-1/8 ln|x-2|+C

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integral of 1/(x(4-x^2))

\int \frac{1}{x ( 4 - x ^{2} )} d x

\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{1}{x ( 4 - x ^{2} )} d x

Take the partial fraction of

\frac{1}{x ( 4 - x ^{2} )} : \frac{1}{4 x} - \frac{1}{8 ( x + 2 )} - \frac{1}{8 ( x - 2 )}

= \int \frac{1}{4 x} - \frac{1}{8 ( x + 2 )} - \frac{1}{8 ( x - 2 )} 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}{8 ( x + 2 )} d x - \int \frac{1}{8 ( x - 2 )} d x

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

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

\int \frac{1}{8 ( x - 2 )} d x = \frac{1}{8} ln ∣ x - 2 ∣

= \frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣ + C

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What is the integral of 1/(x(4-x^2)) ?
The integral of 1/(x(4-x^2)) is 1/4 ln|x|-1/8 ln|x+2|-1/8 ln|x-2|+C