Evaluate $\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$ The task is to evaluate $$\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$$
My best approach has been substitution of $u=x+\frac{1}{2}$, and from there onto (some terrible) trig sub - finally arriving at a messy answer. 
Is there a slicker way of doing this?

We have
\begin{align}
\dfrac{x^2}{\sqrt{1+x+x^2}} & = \dfrac{x^2+x+1}{\sqrt{x^2+x+1}} - \dfrac{x+1}{\sqrt{x^2+x+1}}\\
& = \sqrt{x^2+x+1} - \dfrac12 \dfrac{2x+1}{\sqrt{x^2+x+1}} - \dfrac12 \dfrac1{\sqrt{x^2+x+1}}
\end{align}
