What is the integral of 1/(x^3+x^2+x) ?

The integral of 1/(x^3+x^2+x) 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^3+x^2+x)

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

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 ^{3} + x ^{2} + x} d x

Take the partial fraction of

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

\int x^{2} 1 - x d x

\int_{\frac{π}{4}}^{π} sin (θ) d θ

f (x) = 3 x + x

\frac{d}{d x} (\frac{1}{2} x^{4} - 2 x^{2})

\frac{d}{d x} (ln (\frac{5 x + 1}{x ^{2} - 3}))

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