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?
 A: You have got mixed up with the $\arcsin$. Note that $$\int\frac{1}{\sqrt{\color{red}{1-g^2}}}\, dg = \color{red}{\arcsin g},$$
but
$$\int\frac{1}{\sqrt{\color{blue}{g^2-1}}}\, dg = \color{blue}{\ln\left(\left|g+\sqrt{g^2 -1}\right|\right)}.$$
You can verify this by differentiating the right-hand side, or try to derive this by using a trig. substitution like $g = \sec \theta$. See the bottom two answers here for approaches that don't involve hyperbolic functions. 
A: Here $$\int \frac{1}{\sqrt{g^2-1}} \ dg=...$$
Correct is:
$$\int \frac{1}{\sqrt{1-g^2}} \ dg=
   \arcsin g $$
A: As you did, by completing the square and with $y:=\dfrac2{\sqrt3}\left(x+\dfrac12\right)$,
$$I:=\int\frac{dx}{\sqrt{x^2+x+1}}=\int\frac{dy}{\sqrt{y^2+1}}.$$
Then by some magic $y:=\dfrac12\left(t-\dfrac1t\right)$ yields $dy=\dfrac12\left(1+\dfrac1{t^2}\right)$ and
$$\int\frac{dy}{\sqrt{y^2+1}}=\int\frac{\dfrac12\left(1+\dfrac1{t^2}\right)}{\dfrac12\left(t+\dfrac1{t}\right)}dt=\log t.$$
Now we need to invert,
$$y=\dfrac12\left(t-\dfrac1t\right)\iff t^2-2yt-1=0\iff t=y\pm{\sqrt{y^2+1}}.$$
Finally, choosing the plus sign and ignoring the additive constant,
$$I=\log\left(x+\dfrac12+\sqrt{x^2+x+1}\right).$$
