Integral of this weird function $ \int \frac{1}{(x^2+x+1)^2}dx $

Question

I put this equation into Symbolab and it produced me a very complex result Basically this is the result but I believe there is a simpler way to solve this question

$$ \frac{2}{3\sqrt{3}}\left(2\arctan \left(\frac{2x+1}{\sqrt{3}}\right)+\sin \left(2\arctan \left(\frac{2x+1}{\sqrt{3}}\right)\right)\right)+C $$

This was my question $$ \int \frac{1}{\left(x^2+x+1\right)^2}dx $$

Answer 1

Substitute $y=x+\frac12$\begin{align} &\int \frac{1}{(x^2+x+1)^2}dx\\ =& \int \frac{1}{(y^2+\frac34)^2}dy = \int \frac{2}{3y}\ d\bigg( \frac{y^2}{y^2 +\frac34}\bigg) \overset{ibp}=\frac{2y}{3(y^2+\frac34)}+\frac23\int \frac1{y^2+\frac34}dy\\ =&\ \frac{8y}{3(4y^2+3)}+ \frac4{3\sqrt3}\tan^{-1}\frac{2y}{\sqrt3}+C \end{align}

Answer 2

Under $x+\frac12=\frac{\sqrt3}{2}\tan\theta$ or $\theta=\arctan\bigg(\frac{2x+1}{\sqrt3}\bigg)$, one has \begin{eqnarray} &&\int \frac{1}{(x^2+x+1)^2}dx\\ &=& \int \frac{1}{((x+\frac12)^2+\frac34)^2}dy\\ &=& \frac{8\sqrt3}{9}\int\frac{1}{\sec^4\theta}\sec^2\theta d\theta\\ &=& \frac{8\sqrt3}{9}\int\cos^2\theta d\theta\\ &=& \frac{4\sqrt3}{9}\int(1+\cos(2\theta)) d\theta\\ &=& \frac{2\sqrt3}{9}\bigg[2\theta+\sin(2\theta)\bigg]+C. \end{eqnarray} Using $$ \sin(2x)=\frac{2\tan x}{1+\tan^2x} $$ one has $$ \sin(2\theta) = \frac{2x+1}{2\sqrt3}\frac{1}{x^2+x+1}$$ and hence $$ \int \frac{1}{(x^2+x+1)^2}dx=\frac{2\sqrt3}{9}\bigg[2\arctan\bigg(\frac{2x+1}{\sqrt3}\bigg)+\frac{2x+1}{2\sqrt3}\frac{1}{x^2+x+1}\bigg]+C. $$