How to evaluate $\int\frac{x}{\sqrt{x^2+x+1}}dx\;?$ How do you integrate,
$$\int \frac{x}{\sqrt{x^2+x+1}} dx $$
I am trying to use trig substitution, but I am having trouble finding a perfect square which works.

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{x}{\sqrt{x^2+x+1}}dx$$
Let $$\displaystyle x=\left(y-\frac{1}{2}\right)\;,$$ Then $dx = dy$
So $$\displaystyle x^2+x+1 = \left(y^2+\frac{1}{4}-y\right)+\left(y-\frac{1}{2}\right)+ 1= y^2+\frac{3}{4}$$.
So $$\displaystyle I = \int\frac{\left(y-\frac{1}{2}\right)}{\sqrt{y^2+\frac{3}{4}}}dy = \int\frac{y}{\sqrt{y^2+\frac{3}{4}}}dy-\frac{1}{2}\int \frac{1}{\sqrt{y^2+\frac{3}{4}}}dy$$
So $$\displaystyle I = \sqrt{y^2+\frac{3}{4}}-\frac{1}{2}\cdot \ln \left|y+\sqrt{y^2+\frac{3}{4}}\right|+\mathcal{C}\;,$$ Where $\displaystyle y=\left(x-\frac{1}{2}\right)$
