Help finding integral: $\int \frac{dx}{x\sqrt{1 + x + x^2}}$

Question

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

Answer 1

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

$$\begin{eqnarray*} \int \frac{dx}{x\sqrt{1+x+x^{2}}} &=&\int \frac{2}{t^{2}-1}\,dt \\ &=&-2\operatorname{arctanh}t+C \\ &=&-2\operatorname{arctanh}\left( \sqrt{1+x+x^{2}}-x\right)+C. \end{eqnarray*}$$

Answer 2

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 $x$.

Answer 3

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