<s>[INST] How do I evaluate this integral ? How do you evaluate this integral $\int \frac{x+1}{(x^2-x+1)^3}dx$ ?
I found out:
$\int \frac{x+1}{(x^2-x+1)^3}dx = \frac{1}{2}\int \frac{2x-1}{(x^2-x+1)^3}dx + \int \frac{3}{(x^2-x+1)^3}dx
$
First integral is nicely solvable now. 
I have rearrange the second one:
$
\int \frac{3}{(x^2-x+1)^3}dx = \int \frac{3}{((x-\frac{1}{2})^2+\frac{3}{4})^3}dx=\frac{1}{64}\int \frac{3}{((2x-1)^2+3)^3}dx
$
Let $y=2x-1$, then I have:
$\frac{1}{128}\int \frac{3}{(y^2+3)^3}dy$
What is this integral equeal to ?
 [/INST] At last I have solved it myself. 
$
arctg(y)=\int \frac{1}{y^2+1}dy =
$
Using per parters
$
=\frac{1}{y^2+1}dy-\int\frac{-2y^2}{(y^2+1)^2}dy=\frac{1}{y^2+1}+2\int\frac{y^2+1-1}{(y^2+1)^2}dy=\frac{1}{y^2+1}+2\int\frac{1}{y^2+1}dy-2\int\frac{1}{(y^2+1)2}dy
$
Now I know what is the $\int\frac{1}{(y^2+1)^2}$ equal to.
By proceeding same steps as I did above I will get what I need.
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