<s>[INST] 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.
 [/INST] I will give you a hint:
For second integral take $-1/x^2$ common becase if you take $t= 1/x$ then $dt=-1/x^2dx $ Now substitute t in place of $1/x$ so it will be converted into simple integral.
so next step becomes 
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^3}-\frac{1}{x^2}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}}dx$$
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int-1/x^2\frac{-1/x + 1}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}}dx$$
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{1-t}{\sqrt{1-t+t^2}}dt$$
Now I hope you can solve further easily by method of making into perfecting square and then using the standard formula.Solve the integration and place value of $t=1/x$
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