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

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

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integral of 1/(x^4-x)

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

- ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln x^{2} + x + 1 + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{3 ( x - 1 )} d x = \frac{1}{3} ln ∣ x - 1 ∣

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

= - ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln x^{2} + x + 1

Add a constant to the solution

= - ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln x^{2} + x + 1 + C

\frac{d}{d x} (\frac{2 x ^{2} tan ( x )}{sec ( x )})

\int sec^{2} (θ) tan^{3} (θ) d θ

\int \frac{( x + 2 )}{x} d x

x \to 0 lim (x^{2} ln ∣ x ∣)

\frac{\partial}{\partial x} (y^{x^{y}})

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