<s>[INST] Finding the integral $\int \sqrt{1-\frac{3}{x}+\frac{1}{x^2}} \, dx$ How to find the integral $$\displaystyle\int \sqrt{1-\dfrac{3}{x}+\dfrac{1}{x^2}}\mathrm dx$$ ?
I cannot think of any suitable method. I tried trigonometric substitutions but they were of no use. Hints please!
 [/INST] Use the Euler substitution of the first kind, let $\sqrt{x^2-3x+1}=x+t$,we have
\begin{align*}
\int{\frac{\sqrt{x^2-3x+1}}{x}}\text{d}x&=-2\int{\frac{\left( t^2+3t+1 \right) ^2}{\left( 1-t^2 \right) \left( 2t+3 \right) ^2}}\text{d}t \\
&=-2\int\left (  \frac{1}{2}\frac{1}{t+1}-\frac{3}{2}\frac{1}{2t+3}-\frac{5}{4}\frac{1}{\left (  2t+3\right )^{2}}-\frac{1}{2}\frac{1}{t-1}-\frac{1}{4}\right ) \mathrm{d}t
\end{align*}
then you can take it from here.
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