Help finding integral: $\int \frac{dx}{x\sqrt{1 + x + x^2}}$ 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
 A: Make the substitution $x = \frac{1}{t}$ and this reduces to finding
$$\int \frac{\text{d}t}{\sqrt{t^2 + t + 1}}$$
which can easily be reduced to finding the standard integral:
$$ \int \frac{\text{d}z}{\sqrt{z^2 + 1}} = \sinh^{-1}(z) + C$$
This substitution can be used for finding
$$\int \frac{\text{d}x}{x\sqrt{P(x)}}$$
where $P(x)$ is a quadratic polynomial in $x$.
A: $\displaystyle\int\frac{1}{x\sqrt{x^2+x+1}}dx=\int\frac{1}{x\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}dx$.$\;\;$  Now let $x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan\theta$, $dx=\frac{\sqrt{3}}{2}\sec^{2}\theta d\theta$
to get $\displaystyle\int\frac{1}{(\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2})(\frac{\sqrt{3}}{2}\sec\theta)}\frac{\sqrt{3}}{2}\sec^{2}\theta d\theta=\int\frac{\sec\theta}{\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2}}d\theta=\int\frac{1}{\frac{\sqrt{3}}{2}\sin\theta-\frac{1}{2}\cos\theta}d\theta$
$\;\;=\displaystyle\int\frac{1}{-\cos(\theta+\frac{\pi}{3})}d\theta=-\int\sec\big(\theta+\frac{\pi}{3}\big)d\theta=-\ln\left|\sec\big(\theta+\frac{\pi}{3}\big)+\tan\big(\theta+\frac{\pi}{3}\big)\right|+C$
$\;\;=-\displaystyle\ln\left|\frac{-2\sqrt{x^2+x+1}}{\sqrt{3}x}-\frac{x+2}{\sqrt{3}x}\right|+C=-\ln\left|\frac{2\sqrt{x^2+x+1}+x+2}{\sqrt{3}x}\right|+C$
$\;\;\displaystyle=\ln\left|\frac{\sqrt{3}x}{2\sqrt{x^2+x+1}+x+2}\right|+C=\ln|x|-\ln\left(2\sqrt{x^2+x+1}+x+2\right)+C$
A: Since the integrand is a quadratic irrational function of the type $R(x,\sqrt{1+x+x^{2}})$, you may use the Euler substitution $\sqrt{1+x+x^{2}}=x+t$. You get
$$\begin{eqnarray*}
\int \frac{dx}{x\sqrt{1+x+x^{2}}} &=&\int \frac{2}{t^{2}-1}\,dt \\
&=&-2\operatorname{arctanh}t+C \\
&=&-2\operatorname{arctanh}\left( \sqrt{1+x+x^{2}}-x\right)+C. 
\end{eqnarray*}$$
A: Yes. The substitution $x = \frac{1}{t}$ works. But check for the  minus sign.
It does reduce to integral of $\frac{-dt}{\sqrt{t^2+t+1}}$, which can be reduced further to integral of $\frac{-dz}{z^2 + \frac{\sqrt3}{2}}$
Regards,
Prakash
