Confusing Integral Expression: $\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}$

Question

I have this integral expression to be evaluated:

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

I've followed these steps:

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

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.

Answer 1

Hint

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