A hint to help me integrate $ \int {x^5+1\over x^4+x^3+x^2} \, dx $ Numerator is of a higher degree than the denominator, so after division I get the following:
$$ \int {x^5+1\over x^4+x^3+x^2} \, dx = \frac 12x^2-x + \int {1\over x^2+x+1} \, dx $$

\begin{align}
& \underbrace{x^2+x+1=\left(x^2+x+\frac 1 4 \right)+\frac 3 4}_{\text{completing the square}} = \left( x+\frac 1 2 \right)^2 + \frac 3 4 \\[10pt]
= {} & \frac 3 4 \left( \frac 4 3 \left( x+\frac 1 2 \right)^2 + 1 \right) = \frac 3 4 \left( \left( \frac{2x+1}{\sqrt{3}} \right)^2 +1 \right) = \frac 3 4 (u^2 + 1)
\end{align}
and $du=\dfrac{2}{\sqrt{3}}\,dx$.
And $\displaystyle\int\frac{du}{u^2+1}=\arctan u + C$.
