<s>[INST] Integral of this weird function $ \int \frac{1}{(x^2+x+1)^2}dx $ 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
$$
\int \frac{1}{\left(x^2+x+1\right)^2}dx
$$
 [/INST] 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}
</s>