What is the integral of 1/((x^3-4x)) ?

The integral of 1/((x^3-4x)) is -1/4 ln|x|+1/8 ln|x+2|+1/8 ln|x-2|+C

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integral of 1/((x^3-4x))

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

Factor

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

= \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

= - \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 )}

= - \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 ∣)

Simplify

= - \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

\frac{d}{d θ} (1 + cos (2 θ))

x \to - 1 - lim (\frac{x}{( x + 1 ) ^{2}})

(2 cos (2 x))^{^{'}}

3 s + 2

x \to \infty lim (e^{x} - 3)

What is the integral of 1/((x^3-4x)) ?
The integral of 1/((x^3-4x)) is -1/4 ln|x|+1/8 ln|x+2|+1/8 ln|x-2|+C