Integrating $\int \frac{\sqrt{x^2-x+1}}{x^2}dx$ Evaluate $$I=\int\frac{\sqrt{x^2-x+1}}{x^2}dx$$  I first Rationalized the numerator and got as
$$I=\int\frac{(x^2-x+1)dx}{x^2\sqrt{x^2-x+1}}$$ and splitting we get
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^2}-\frac{1}{x}}{\sqrt{x^2-x+1}}dx$$ i.e.,
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^3}-\frac{1}{x^2}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}}dx$$
First Integral can be evaluated using standard integral. But second one i am not able to do since numerator is not differential of expression inside square root in denominator.

So the last step should be changing variable $t=\frac{2}{\sqrt3}(x-\frac{1}{2})$ and use the identity 
$$\int \frac{1}{\sqrt{x^2+1}}dx=arcsinh(x)$$
so we get
$$I=-\frac{\sqrt{x^2-x+1}}{x}+\frac{3}{2}\int\frac{1}{\sqrt{x^2-x+1}}dx=
 -\frac{\sqrt{x^2-x+1}}{x}+\frac{3}{2}arcsinh[\frac{2}{\sqrt3}(x-\frac{1}{2})]$$
