TITLE: Help finding integral: $\int \frac{dx}{x\sqrt{1 + x + x^2}}$
QUESTION [7 upvotes]: Could someone help me with finding this integral
$$\int  \frac{dx}{x\sqrt{1 + x + x^2}}$$
or give a hint on how to solve it.
Thanks in advance

REPLY [2 votes]: $\displaystyle\int\frac{1}{x\sqrt{x^2+x+1}}dx=\int\frac{1}{x\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}dx$.$\;\;$  Now let $x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan\theta$, $dx=\frac{\sqrt{3}}{2}\sec^{2}\theta d\theta$
to get $\displaystyle\int\frac{1}{(\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2})(\frac{\sqrt{3}}{2}\sec\theta)}\frac{\sqrt{3}}{2}\sec^{2}\theta d\theta=\int\frac{\sec\theta}{\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2}}d\theta=\int\frac{1}{\frac{\sqrt{3}}{2}\sin\theta-\frac{1}{2}\cos\theta}d\theta$
$\;\;=\displaystyle\int\frac{1}{-\cos(\theta+\frac{\pi}{3})}d\theta=-\int\sec\big(\theta+\frac{\pi}{3}\big)d\theta=-\ln\left|\sec\big(\theta+\frac{\pi}{3}\big)+\tan\big(\theta+\frac{\pi}{3}\big)\right|+C$
$\;\;=-\displaystyle\ln\left|\frac{-2\sqrt{x^2+x+1}}{\sqrt{3}x}-\frac{x+2}{\sqrt{3}x}\right|+C=-\ln\left|\frac{2\sqrt{x^2+x+1}+x+2}{\sqrt{3}x}\right|+C$
$\;\;\displaystyle=\ln\left|\frac{\sqrt{3}x}{2\sqrt{x^2+x+1}+x+2}\right|+C=\ln|x|-\ln\left(2\sqrt{x^2+x+1}+x+2\right)+C$