<s>[INST] Evaluation of $\int\frac{dx}{x+ \sqrt{x^2-x+1}}$ Evaluate :
$$\frac{dx}{x+ \sqrt{x^2-x+1}}$$
After dividing and multiply with $x-\sqrt{x^2-x+1}$, I get
$x+\ln |x-1|-\int \frac{\sqrt{x^2-x+1}}{x-1}dx$.
Is $\int \frac{\sqrt{x^2-x+1}}{x-1}dx$ is integrable in terms of elementary functions?
 [/INST] There is a substitution, called the Euler substitution, which solves your integral rather neatly. In your case we set $t = \sqrt{x^2 - x + 1} + x$; after subtracting $x$ from both sides and squaring, we get $x^2 -x + 1 = t^2 - 2xt + x^2$, which can be solved (after subtracting $x^2$ from both sides) to get $x = \frac{t^2 - 1}{2t - 1}$. In this case, $\frac{dt}{dx} = \frac{1}{\sqrt{x^2 - x + 1}}(2x-1) + 1$, and so 
\begin{eqnarray} \int \frac{dx}{x + \sqrt{x^2 - x + 1}} = \int (\frac{1}{x + \sqrt{x^2 - x + 1}}\cdot\frac{1}{\frac{dt}{dx}})\frac{dt}{dx}dx \\= \int \frac{1}{t}\cdot\frac{1}{[\frac{1}{t - (t^2-1)/(2t-1)}\cdot(2(\frac{t^2-1}{2t-1}) - 1)] + 1} dt = \dots \\ \dots = \int\frac{t^2 - t + 1}{3t^3 - 3t^2} dt\end{eqnarray}
which is quite easy to solve directly by partial fractions.
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