How to integrate $\frac{1}{\sqrt{x^2+x+1}}$ 
How to integrate $$\frac{1}{\sqrt{x^2+x+1}}$$

I tried to solve this integral as follows
$\displaystyle
  \int \frac{1}{\sqrt{x^2+x+1}} \ dx=
  \int \frac{1}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}} \ dx=
  \int \frac{1}{\sqrt{(\frac{2x+1}{2})^2+\frac{3}{4}}} \ dx=
  \int \frac{1}{\sqrt{\frac{3}{4}((\frac{2x+1}{\sqrt{3}})^2+1)}} \ dx=
  \int \frac{1}{\sqrt{(\frac{2x+1}{\sqrt{3}})^2+1}}\frac{2}{\sqrt{3}} \ dx$ 
Substitution $t=\frac{2x+1}{\sqrt{3}} ;dt=\frac{2}{\sqrt{3}} \ dx$
$\displaystyle\int \frac{1}{\sqrt{(\frac{2x+1}{\sqrt{3}})^2+1}}\frac{2}{\sqrt{3}} \ dx=
   \int \frac{1}{\sqrt{t^2+1}} \ dt$
Substitution $\sqrt{u-1}= t;\frac{1}{2\sqrt{u-1}} \ du= dt$
$\displaystyle
   \int \frac{1}{\sqrt{t^2+1}} \ dt=
   \int \frac{1}{2 \sqrt{u(u-1)}} \ du=
   \frac{1}{2}\int \frac{1}{\sqrt{u^2-u}} \ du=
   \frac{1}{2}\int \frac{1}{\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}} \ du=
   \int \frac{1}{\sqrt{(2u-1)^2-1}} \ du$
Substitution $g= 2u-1;dg= 2 \ du$
$\displaystyle
   \int \frac{1}{\sqrt{(2u-1)^2-1}} \ du=
   \frac{1}{2}\int \frac{2}{\sqrt{(2u-1)^2-1}} \ du=
   \frac{1}{2}\int \frac{1}{\sqrt{g^2-1}} \ dg=
   \frac{1}{2}\arcsin g +C=\frac{1}{2}\arcsin (2u-1) +C= \frac{1}{2}\arcsin (2(t^2+1)-1) +C=\frac{1}{2}\arcsin (2((\frac{2x+1}{\sqrt{3}})^2+1)-1) +C$
However when I tried to graph it using desmos there was no result, and when i used https://www.integral-calculator.com/ on thi problem it got the result $$\ln\left(\left|2\left(\sqrt{x^2+x+1}+x\right)+1\right|\right)$$
where have I made a mistake?

Here $$\int \frac{1}{\sqrt{g^2-1}} \ dg=...$$
Correct is:
$$\int \frac{1}{\sqrt{1-g^2}} \ dg=
   \arcsin g $$
