Integration: $\int\frac{1}{(x^2+x+1)^{1/2}} dx$ Find the value of $$\int\frac{1}{(x^2+x+1)^{1/2}} dx$$ 
Anyone can provide hint on how to integrate this, and how you know what method to use? (I mean, is there any general guideline to follow for solving?)
Thank you!
 A: If you make the Euler substitution $t=x+\sqrt{x^2+x+1}$, then $\displaystyle x=\frac{t^2-1}{2t+1}$ and $\displaystyle dx=\frac{2(t^2+t+1)}{(2t+1)^2}dt$;
so $\displaystyle\int\frac{1}{\sqrt{x^2+x+1}}dx=\int\frac{1}{t-\frac{t^2-1}{2t+1}}\cdot\frac{2(t^2+t+1)}{(2t+1)^2}dt=\int\frac{2}{2t+1}dt=\ln\lvert2t+1\lvert+C$
$\displaystyle=\ln\left(2x+2\sqrt{x^2+x+1}+1\right)+C$
A: Hint: Notice that  $$x^2 + x + 1 = \Big(x  +\frac{1}{2}\Big)^{2}  +\frac{3}{4} $$
Use $x + \frac{1}{2}= \frac{\sqrt 3}{2} \sinh t $.
A: $\bf{My\; Solution:: }$ Let $\displaystyle I = \int\frac{1}{\sqrt{x^2+x+1}}dx = \int\frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}dx$
Now Let $\displaystyle \left(x+\frac{1}{2}\right) = t\;,$ Then $dx = dt$ and $\displaystyle \frac{\sqrt{3}}{2} = a>0$
So Integral $$\displaystyle I = \int\frac{1}{\sqrt{t^2+a^2}}dt$$
Now Let $t^2+a^2 = y^2\;,$ Then $$\displaystyle tdt = ydy \Rightarrow \frac{dt}{y} = \frac{dy}{t}=\frac{d(t+y)}{(t+y)}$$
So Integral $$\displaystyle I = \int\frac{dt}{y} = \int\frac{d(t+y)}{(t+y)}= \ln \left|t+y\right|+\mathcal{C}$$
So $$\displaystyle I = \int\frac{1}{\sqrt{x^2+x+1}}dx = \ln \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|+\mathcal{C}.$$
A: $$\int\frac1{\sqrt{x^2+x+1}}=\int\frac1{\sqrt{(x+1/2)^2+3/4}}=\ln|x+1/2+\sqrt{x^2+x+1}|+{\rm\color{grey}{ constant}}$$
