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

x \to 0 lim (\frac{( 2 - x ) ^{2} - 4}{x})

x^{3} - x, 3 x

\frac{\partial}{\partial x} (x^{2} sin^{2} (y))

\frac{\partial}{\partial x} (y^{3} sin (4 x))

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

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