Confusing Integral Expression: $\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}$ I have this integral expression to be evaluated:

$$\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}$$

I've followed these steps:


*

*Completed the square in $x$ and then substituted $u=x-\frac{1}{2}.$

*Trig substituted $u = \frac{\sqrt{3}}{2}\tan t.$ and then simplified the equation thereby giving: 
$$\int{\frac{\sqrt{3}\sec^2t}{\sqrt{3}\tan t+\sqrt{3}\sec t+1}dt}$$ 
I know up to this, but I'm unclear what to do next.
Feel free to correct me if I'm wrong.

Hint
Start considering the integrand as $$\frac{1}{x+\sqrt{x^2-x+1}}=\frac{1}{x+\sqrt{x^2-x+1}}\times\frac{x-\sqrt{x^2-x+1}}{x-\sqrt{x^2-x+1}}$$
$$\frac{1}{x+\sqrt{x^2-x+1}}=\frac{x-\sqrt{x^2-x+1}}{x-1}=\frac x {x-1}-\frac{\sqrt{x^2-x+1}}{x-1}$$
