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

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

8 sin (8 x - 1)

\frac{d}{d x} (e^{x} (2 x + 3)^{3})

\frac{d ^{2}}{d x ^{2}} (e^{\frac{3}{x}})

\int \frac{9 x}{x ^{2} + 1} d x

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