<s>[INST] Solve $\int\frac{2x-3}{(x^2+x+1)^2}dx$ $\int\frac{2x-3}{(x^2+x+1)^2}dx$

$\int\frac{2x-3}{(x^2+x+1)^2}dx=\int\frac{2x+1}{(x^2+x+1)^2}dx-\int\frac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
 [/INST] Hint:
As $x^2+x+1=\dfrac{(2x+1)^2+3}4,$  set $2x+1=\sqrt3\tan t$
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