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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{2}{4} x^{- 8} d x

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

\int \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

\int (ln (\frac{3 a}{x}))^{2} d x

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