Evaluate: $\int \frac{dx}{(x^2+x+1)^2},$ without using any kind of substitutions. I know there exists a reduction formula for the integral: $$\int\frac1{(ax^2+bx+c)^n}\,dx.$$ But this uses substitutions. So for the integral in question, let:
\begin{align}
I&=\int\frac{dx}{(x^2+x+1)^2}\\\\
&=\int\frac{dx}{\{(x-\omega)(x-\omega^2)\}^2}\\\\
&=\int\frac{dx}{(x-\omega)^2(x-\omega^2)^2},
\end{align}
Where, $\omega$ is one of the complex cube roots of unity. According to me, resolving $\dfrac1{(x-\omega)^2(x-\omega^2)^2}$ into partial fractions, would serve our purpose. Is this a proper way to approach? Thanks in advance. 

\begin{align}
I=\int \frac{dx}{(x^2+x+1)^2}&=\frac{1}{B}\int \frac{1}{2x+1}d(A-\frac{B}{x^2+x+1}) \\
&=\frac{1}{B}\left[ \frac{1}{2x+1}(A-\frac{B}{x^2+x+1})+\int (A-\frac{B}{x^2+x+1})\frac{2}{(2x+1)^2}dx \right] \\
\end{align}
Note
\begin{align}
&(A-\frac{B}{x^2+x+1})\frac{2}{(2x+1)^2} \\
&=\frac{Ax^2+Ax+A-B}{x^2+x+1} \frac{2}{4x^2+4x+1}
\end{align}
So,we can let
$$ A=4,A-B=1\Rightarrow A=4,B=3 $$
Thus
\begin{align}
I&=\frac{1}{3}\left[ \frac{2x+1}{x^2+x+1}+2\int \frac{1}{x^2+x+1} dx \right] \\
&=\frac{1}{3}\left[ \frac{2x+1}{x^2+x+1}+2\int \frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}} dx \right] \\
&=\frac{1}{3}\left[ \frac{2x+1}{x^2+x+1}+\frac{4}{\sqrt{3}}\arctan \frac{2x+1}{\sqrt{3}} \right]+C \\
\end{align}
