I've tried to use substitution for indefinite integral:
$$\int{\frac{dx}{x + \sqrt{x^{2} + x + 1}}}$$
using Euler substitution $t = \sqrt{x^2 + x + 1} - x$, $x = \frac{1 - t^2}{2t - 1}$
then: $$ \begin{align} dx =& \frac{dt}{(\sqrt{x^2 + x + 1} - x)'} = \\ =& \frac{dt}{\frac{2x + 1 - 2\sqrt{x^2 + x + 1}}{2\sqrt{x^2 + x + 1}}} = \\ =& \frac{2\sqrt{x^2 + x + 1}}{2x + 1 - 2\sqrt{x^2 + x + 1}}dt = \\ =& \frac{2(\sqrt{x^2 + x + 1} - x + x)}{2x + 1 - 2(\sqrt{x^2 + x + 1} - x + x)}dt = \\ =& \frac{2(t + x)}{2x + 1 - 2(t + x)}dt = \\ =& \frac{2(t + x)}{1 - 2t}dt \end{align} $$
then apply to integral:
$$ \begin{align} \int{\frac{\frac{2(t + x)}{1 - 2t}}{\frac{2 - 2t^{2}}{2t - 1} + t}dt} =& \int{\frac{\frac{2(t + x)}{(-1)(2t - 1)}}{\frac{2 - t}{2t - 1}}dt} = \\ =& \int{\frac{2t + 2x}{t - 2}dt} \\ =& \int{\frac{2t + \frac{2 - 2t^{2}}{2t - 1}}{t - 2}dt} \\ =& \int{\frac{\frac{2t^{2} - 2t + 2}{2t - 1}}{t - 2}dt} \\ =& \int{\frac{2t^{2} - 2t + 2}{2t^{2} - 5t + 2}dt} \end{align} $$
However when I checked with WolframAlpha I got different results for both integrals so I assume I've calculated incorrectly this substitution, but not sure where I am wrong.
Thanks