Integration: $\int\frac{1}{(x^2+x+1)^{1/2}} dx$

Question

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!

Answer 1

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 $.

Answer 2

$$\int\frac1{\sqrt{x^2+x+1}}=\int\frac1{\sqrt{(x+1/2)^2+3/4}}=\ln|x+1/2+\sqrt{x^2+x+1}|+{\rm\color{grey}{ constant}}$$

Answer 3

$\bf{My\; Solution:: }$ 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 $$\displaystyle I = \int\frac{1}{\sqrt{x^2+x+1}}dx = \ln \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|+\mathcal{C}.$$

Answer 4

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 $\displaystyle\int\frac{1}{\sqrt{x^2+x+1}}dx=\int\frac{1}{t-\frac{t^2-1}{2t+1}}\cdot\frac{2(t^2+t+1)}{(2t+1)^2}dt=\int\frac{2}{2t+1}dt=\ln\lvert2t+1\lvert+C$

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