<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] Using Integration by Parts, We get
Let $$I =\int \sqrt{x^2-x+1}\cdot \frac{1}{x^2}dx = -\frac{\sqrt{x^2-x+1}}{x}+\int\frac{2x-1}{2x\sqrt{x^2-x+1}}dx$$
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So we get $$I = -\frac{\sqrt{x^2-x+1}}{x}+\int\frac{1}{\sqrt{x^2-x+1}}dx-\frac{1}{2}\int\frac{1}{x\sqrt{x^2-x+1}}dx$$
Now for last Integral, Let $$J = \int \frac{1}{x\sqrt{x^2-x+1}}dx$$
Put $\displaystyle x=\frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So we get $$J=-\int\frac{1}{\sqrt{t^2-t+1}}dt$$
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