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

Notice

$$ x^2 - x + 1 = \left(x-\frac{1}{2}\right)^2 + 1 - \frac{1}{4} = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4}$$

Completing the square will yield $$ x^2 - x + 1 = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4} $$ Normally, we will let $u=x-\frac{1}{2}$. However it can also be solved by letting $x-\frac{1}{2}=\frac{\sqrt3}{2}\sinh t$ and $dx=\frac{\sqrt3}{2}\cosh t\ dt$ which yields $$ \begin{align} \int \frac{dx}{\sqrt{x^{2}-x+1}}&=\int \frac{\frac{\sqrt3}{2}\cosh t\ dt}{\sqrt{\frac{3}{4}\sinh^2 t+\frac{3}{4}}}\\ &=\int \frac{\cosh t\ dt}{\sqrt{\cosh^2 t}}\\ &=\int \ dt\\ &=t+C \end{align} $$ where $\sinh t=\dfrac{2x-1}{\sqrt3}\;\Rightarrow\; t=\sinh^{-1}\left(\dfrac{2x-1}{\sqrt3}\right)$. Thus $$ \int \frac{dx}{\sqrt{x^{2}-x+1}}=\sinh^{-1}\left(\dfrac{2x-1}{\sqrt3}\right)+C. $$ As your book's solution.

Note that these are actually the same answer, since:

$$ \sinh^{-1} x = \ln \left(x + \sqrt{x^2+1} \right) $$

and

$$ \frac{2x-1}{\sqrt{3}} + \sqrt{\left(\frac{2x-1}{\sqrt{3}}\right)^2 + 1} = \frac{2x - 1 + 2\sqrt{x^2 - x + 1}}{\sqrt{3}}$$

So the answer you got is also correct.