<s>[INST] Integral of $\int \frac{x+1}{(x^2+x+1)^2}dx$ I'm currently learning Calculus II and I have the following integral:
Integal of $\large{\int \frac{x+1}{(x^2+x+1)^2}dx}$ 
I've tried with partial fractions but it led nowhere, I've tried with substitution, but I failed again.
If it helps, I know the answer, but I don't know how to get to it.
Answer: $\large{\frac{1}{3}\frac{x-1}{x^2+x+1}} + \frac{1}{3}2\sqrt{3} \arctan{2x+1\sqrt{3}} + C$
 [/INST] Hint:
\begin{align*}
\int\frac{x+1}{(x^2+x+1)^2}dx&=\frac{1}{2}\int{\frac{2x+1}{(x^2+x+1)^2}}dx+\frac{1}{2}\int\frac{1}{(x^2+x+1)^2}dx\\
&=\frac{1}{2}\int\frac{d(x^2+x+1)}{(x^2+x+1)^2}+\frac{1}{2}\int{\frac{1}{\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2}}dx
\end{align*}
The last of these integrals can be found using the trigonometric sustitution $x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan t$.
</s>