Integral: $\int \frac{dx}{\sqrt{x^{2}-x+1}}$ How do I integrate this? $$\int \frac{dx}{\sqrt{x^{2}-x+1}}$$
I tried solving it, and I came up with $\ln\left | \frac{2\sqrt{x^{2}-x+1}+2x-1}{\sqrt{3}} \right |+C$. But the answer key says that the answer should be $\sinh^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right )+C$. Any answer is very appreciated.

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.
