What is the integral of 1/((x(x^2+x+1))) ?

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

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integral of 1/((x(x^2+x+1)))

integral

\frac{1}{( x ( x ^{2} + x + 1 ) )}

ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{- x - 1}{x ^{2} + x + 1} d x = - \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1))

= ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1))

Add a constant to the solution

= ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1)) + C

5 (1 - x)^{4}

\frac{1}{( 1 - x ^{3} )}

y = 2 x^{3} - 5 x + 4

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

\frac{s}{( s ^{2} + 4 )}

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