What is the integral of 1/(x(x^4+x^2+1)^2) ?

The integral of 1/(x(x^4+x^2+1)^2) is

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integral of 1/(x(x^4+x^2+1)^2)

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

- \frac{1}{4} ln 4 x^{4} + 4 x^{2} + 4 - \frac{1}{6 3} arctan (\frac{1}{3} (2 x^{2} + 1)) + \frac{1}{6 3} sin (2 arctan (\frac{1}{3} (2 x^{2} + 1))) - \frac{2}{3 3} arctan (\frac{2 x ^{2} + 1}{3}) - \frac{1}{3 3} sin (2 arctan (\frac{2 x ^{2} + 1}{3})) + ln (x) + \frac{1}{4 ( x ^{4} + x ^{2} + 1 )} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= \frac{1}{2} \cdot \int \frac{1}{u} + \frac{- u - 1}{u ^{2} + u + 1} + \frac{- u - 1}{( u ^{2} + u + 1 ) ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int \frac{- u - 1}{( u ^{2} + u + 1 ) ^{2}} d u = - 8 (- \frac{1}{4 ( 4 u ^{2} + 4 u + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 u + 1)) + sin (2 arctan (\frac{1}{3} (2 u + 1))))) - \frac{2}{3 3} (2 arctan (\frac{2 u + 1}{3}) + sin (2 arctan (\frac{2 u + 1}{3})))

= \frac{1}{2} (ln ∣ u ∣ - \frac{1}{2} ln 4 u^{2} + 4 u + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 u + 1)) - 8 (- \frac{1}{4 ( 4 u ^{2} + 4 u + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 u + 1)) + sin (2 arctan (\frac{1}{3} (2 u + 1))))) - \frac{2}{3 3} (2 arctan (\frac{2 u + 1}{3}) + sin (2 arctan (\frac{2 u + 1}{3}))))

Substitute back

u = x^{2}

= \frac{1}{2} ln x^{2} - \frac{1}{2} ln 4 (x^{2})^{2} + 4 x^{2} + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x^{2} + 1)) - 8 - \frac{1}{4 ( 4 ( x ^{2} ) ^{2} + 4 x ^{2} + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x^{2} + 1)) + sin (2 arctan (\frac{1}{3} (2 x^{2} + 1)))) - \frac{2}{3 3} (2 arctan (\frac{2 x ^{2} + 1}{3}) + sin (2 arctan (\frac{2 x ^{2} + 1}{3})))

Simplify

\frac{1}{2} ln x^{2} - \frac{1}{2} ln 4 (x^{2})^{2} + 4 x^{2} + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x^{2} + 1)) - 8 - \frac{1}{4 ( 4 ( x ^{2} ) ^{2} + 4 x ^{2} + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x^{2} + 1)) + sin (2 arctan (\frac{1}{3} (2 x^{2} + 1)))) - \frac{2}{3 3} (2 arctan (\frac{2 x ^{2} + 1}{3}) + sin (2 arctan (\frac{2 x ^{2} + 1}{3}))) : - \frac{1}{4} ln 4 x^{4} + 4 x^{2} + 4 - \frac{1}{6 3} arctan (\frac{1}{3} (2 x^{2} + 1)) + \frac{1}{6 3} sin (2 arctan (\frac{1}{3} (2 x^{2} + 1))) - \frac{2}{3 3} arctan (\frac{2 x ^{2} + 1}{3}) - \frac{1}{3 3} sin (2 arctan (\frac{2 x ^{2} + 1}{3})) + ln (x) + \frac{1}{4 ( x ^{4} + x ^{2} + 1 )}

= - \frac{1}{4} ln 4 x^{4} + 4 x^{2} + 4 - \frac{1}{6 3} arctan (\frac{1}{3} (2 x^{2} + 1)) + \frac{1}{6 3} sin (2 arctan (\frac{1}{3} (2 x^{2} + 1))) - \frac{2}{3 3} arctan (\frac{2 x ^{2} + 1}{3}) - \frac{1}{3 3} sin (2 arctan (\frac{2 x ^{2} + 1}{3})) + ln (x) + \frac{1}{4 ( x ^{4} + x ^{2} + 1 )}

Add a constant to the solution

= - \frac{1}{4} ln 4 x^{4} + 4 x^{2} + 4 - \frac{1}{6 3} arctan (\frac{1}{3} (2 x^{2} + 1)) + \frac{1}{6 3} sin (2 arctan (\frac{1}{3} (2 x^{2} + 1))) - \frac{2}{3 3} arctan (\frac{2 x ^{2} + 1}{3}) - \frac{1}{3 3} sin (2 arctan (\frac{2 x ^{2} + 1}{3})) + ln (x) + \frac{1}{4 ( x ^{4} + x ^{2} + 1 )} + C

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What is the integral of 1/(x(x^4+x^2+1)^2) ?
The integral of 1/(x(x^4+x^2+1)^2) is