How do I evaluate this integral :$\int \frac{\sqrt{-x^2-x+2}}{x^2}dx$? Is there someone who can show me how to evaluate this integral:
$$\int \frac{\sqrt{-x^2-x+2}}{x^2}dx.$$
I have tried many changes of variables but I haven't succeeded yet. Thank you for any help.

$\begin{gathered}
  \int {\frac{{\sqrt { - {x^2} - x + 2} }}{{{x^2}}}} \,dx \hfill \\
   =  - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \int {\frac{{\left( { - 2x - 1} \right)}}{{2\sqrt { - {x^2} - x + 2} }}} \frac{{ - 1}}{x}\,dx\quad \quad \left( {{\text{Integrating by parts}}} \right) \hfill \\
   =  - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \int {\frac{{2x + 1}}{{2x\sqrt { - {x^2} - x + 2} }}} \,dx \hfill \\
   =  - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \left[ {\int {\frac{{dx}}{{\sqrt {\left( {2 + x} \right)\left( {1 - x} \right)} }} + \int {\frac{{dx}}{{2x\sqrt { - {x^2} - x + 2} }}} } } \right] \hfill \\
   =  - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \left[ {2{{\sin }^{ - 1}}\sqrt {\frac{{x + 2}}{3}}  - \frac{1}{{2\sqrt 2 }}\log \left| {\frac{1}{{x\sqrt 2 }} - \frac{1}{{4\sqrt 2 }} + \sqrt {\frac{1}{{2{x^2}}} - \frac{1}{{4x}} - \frac{1}{4}} } \right|} \right]+c \hfill \\ 
\end{gathered} $
