What is the integral of 8/(x^3-4x) ?

The integral of 8/(x^3-4x) is -2ln|x|+ln|x+2|+ln|x-2|+C

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integral of 8/(x^3-4x)

\int \frac{8}{x ^{3} - 4 x} d x

- 2 ln ∣ x ∣ + ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{8}{x ^{3} - 4 x} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{3} - 4 x} d x

Factor

x^{3} - 4 x : - (- x^{3} + 4 x)

= 8 \cdot \int \frac{1}{- ( - x ^{3} + 4 x )} d x

Take the constant out:

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

= 8 (- \int \frac{1}{- x ^{3} + 4 x} d x)

Take the partial fraction of

\frac{1}{- x ^{3} + 4 x} : \frac{1}{4 x} - \frac{1}{8 ( x + 2 )} - \frac{1}{8 ( x - 2 )}

= 8 (- \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

= 8 (- (\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 ∣

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

Simplify

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

= - 2 ln ∣ x ∣ + ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣

Add a constant to the solution

= - 2 ln ∣ x ∣ + ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

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What is the integral of 8/(x^3-4x) ?
The integral of 8/(x^3-4x) is -2ln|x|+ln|x+2|+ln|x-2|+C