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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{x} + \frac{1}{2 x + 1} + \frac{1}{2 x - 1} 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}{x} d x + \int \frac{1}{2 x + 1} d x + \int \frac{1}{2 x - 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

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